CONVERSION FACTORS FROM ENGLISH TO SI UNITS
Length:
Area:
Volume:
Force:
1 ft
1 ft
1 ft
1 in.
1 in.
1 in.
2
5
5
5
5
5
5
0.3048 m
30.48 cm
304.8 mm
0.0254 m
2.54 cm
25.4 mm
24
Stress:
1 lb>ft 2
1 lb>ft 2
1 U.S. ton>ft 2
1 kip>ft 2
1 lb>in2
5
5
5
5
5
Unit weight:
1 lb>ft 3
1 lb>in3
5 0.1572 kN>m3
5 271.43 kN>m3
Moment:
1 lb-ft
1 lb-in.
5 1.3558 N # m
5 0.11298 N # m
Energy:
1 ft-lb
5 1.3558 J
Moment of
inertia:
1 in4
1 in4
5 0.4162 3 106 mm4
5 0.4162 3 1026 m4
Section
modulus:
1 in3
1 in3
5 0.16387 3 105 mm3
5 0.16387 3 1024 m3
Hydraulic
conductivity:
1 ft>min
1 ft>min
1 ft>min
1 ft>sec
1 ft>sec
1 in.>min
1 in.>sec
1 in.>sec
5 0.3048 m>min
5 30.48 cm>min
5 304.8 mm>min
5 0.3048 m>sec
5 304.8 mm>sec
5 0.0254 m>min
5 2.54 cm>sec
5 25.4 mm>sec
2
1 ft
1 ft 2
1 ft 2
1 in2
1 in2
1 in2
5
5
5
5
5
5
929.03 3 10 m
929.03 cm2
929.03 3 102 mm2
6.452 3 1024 m2
6.452 cm2
645.16 mm2
1 ft 3
1 ft 3
1 in3
1 in3
5 28.317 3 1023 m3
5 28.317 3 103 cm3
5 16.387 3 1026 m3
5 16.387 cm3
1 lb
1 lb
1 lb
1 kip
1 U.S. ton
1 lb
1 lb>ft
5 4.448 N
5 4.448 3 1023 kN
5 0.4536 kgf
5 4.448 kN
5 8.896 kN
5 0.4536 3 1023 metric ton
5 14.593 N>m
Coefficient of
consolidation:
1 in2>sec
1 in2>sec
1 ft 2>sec
47.88 N>m2
0.04788 kN>m2
95.76 kN>m2
47.88 kN>m2
6.895 kN>m2
5 6.452 cm2>sec
5 20.346 3 103 m2>yr
5 929.03 cm2>sec
Principles of
Foundation Engineering, SI
Seventh Edition
BRAJA M. DAS
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
Principles of Foundation Engineering, SI
Seventh Edition
Author Braja M. Das
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Christopher M. Shortt
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Contents
Preface xvii
1
Geotechnical Properties of Soil 1
1.1 Introduction 1
1.2 Grain-Size Distribution 2
1.3 Size Limits for Soils 5
1.4 Weight–Volume Relationships 5
1.5 Relative Density 10
1.6 Atterberg Limits 15
1.7 Liquidity Index 16
1.8 Activity 17
1.9 Soil Classification Systems 17
1.10 Hydraulic Conductivity of Soil 25
1.11 Steady-State Seepage 28
1.12 Effective Stress 30
1.13 Consolidation 32
1.14 Calculation of Primary Consolidation Settlement 37
1.15 Time Rate of Consolidation 38
1.16 Degree of Consolidation Under Ramp Loading 44
1.17 Shear Strength 47
1.18 Unconfined Compression Test 52
1.19 Comments on Friction Angle, fr 54
1.20 Correlations for Undrained Shear Strength, Cu 57
1.21 Sensitivity 57
Problems 58
References 62
vii
viii
Contents
2
Natural Soil Deposits and Subsoil Exploration 64
2.1 Introduction 64
Natural Soil Deposits
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
64
Soil Origin 64
Residual Soil 66
Gravity Transported Soil 67
Alluvial Deposits 68
Lacustrine Deposits 70
Glacial Deposits 70
Aeolian Soil Deposits 71
Organic Soil 73
Some Local Terms for Soils 73
Subsurface Exploration
74
2.11 Purpose of Subsurface Exploration 74
2.12 Subsurface Exploration Program 74
2.13 Exploratory Borings in the Field 77
2.14 Procedures for Sampling Soil 81
2.15 Split-Spoon Sampling 81
2.16 Sampling with a Scraper Bucket 89
2.17 Sampling with a Thin-Walled Tube 90
2.18 Sampling with a Piston Sampler 92
2.19 Observation of Water Tables 92
2.20 Vane Shear Test 94
2.21 Cone Penetration Test 98
2.22 Pressuremeter Test (PMT) 107
2.23 Dilatometer Test 110
2.24 Coring of Rocks 113
2.25 Preparation of Boring Logs 117
2.26 Geophysical Exploration 118
2.27 Subsoil Exploration Report 126
Problems 126
References 130
3
Shallow Foundations: Ultimate Bearing Capacity 133
3.1
3.2
3.3
3.4
Introduction 133
General Concept 133
Terzaghi’s Bearing Capacity Theory 136
Factor of Safety 140
Contents
3.5
3.6
3.7
3.8
3.9
3.10
Modification of Bearing Capacity Equations for Water Table 142
The General Bearing Capacity Equation 143
Case Studies on Ultimate Bearing Capacity 148
Effect of Soil Compressibility 153
Eccentrically Loaded Foundations 157
Ultimate Bearing Capacity under Eccentric Loading—One-Way
Eccentricity 159
3.11 Bearing Capacity—Two-way Eccentricity 165
3.12 Bearing Capacity of a Continuous Foundation Subjected to
Eccentric Inclined Loading 173
Problems 177
References 179
4
Ultimate Bearing Capacity of Shallow Foundations:
Special Cases 181
4.1 Introduction 181
4.2 Foundation Supported by a Soil with a Rigid Base at Shallow
Depth 181
4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain
by Weaker Soil 190
4.4 Bearing Capacity of Layered Soil: Weaker Soil Underlain
by Stronger Soil 198
4.5 Closely Spaced Foundations—Effect on Ultimate Bearing
Capacity 200
4.6 Bearing Capacity of Foundations on Top of a Slope 203
4.7 Seismic Bearing Capacity of a Foundation at the Edge
of a Granular Soil Slope 209
4.8 Bearing Capacity of Foundations on a Slope 210
4.9 Foundations on Rock 212
4.10 Uplift Capacity of Foundations 213
Problems 219
References 221
5
Shallow Foundations: Allowable Bearing Capacity
and Settlement 223
5.1 Introduction 223
Vertical Stress Increase in a Soil Mass Caused by Foundation Load
5.2 Stress Due to a Concentrated Load 224
224
ix
x Contents
5.3 Stress Due to a Circularly Loaded Area 224
5.4 Stress below a Rectangular Area 226
5.5 Average Vertical Stress Increase Due to a Rectangularly
Loaded Area 232
5.6 Stress Increase under an Embankment 236
5.7 Westergaard’s Solution for Vertical Stress Due to a
Point Load 240
5.8 Stress Distribution for Westergaard Material 241
Elastic Settlement 243
Elastic Settlement of Foundations on Saturated Clay (␮S ⫽ 0.5) 243
Settlement Based on the Theory of Elasticity 245
Improved Equation for Elastic Settlement 254
Settlement of Sandy Soil: Use of Strain Influence Factor 258
Settlement of Foundation on Sand Based on Standard Penetration
Resistance 263
5.14 Settlement in Granular Soil Based on Pressuremeter Test
(PMT) 267
5.9
5.10
5.11
5.12
5.13
Consolidation Settlement
273
5.15 Primary Consolidation Settlement Relationships 273
5.16 Three-Dimensional Effect on Primary Consolidation
Settlement 274
5.17 Settlement Due to Secondary Consolidation 278
5.18 Field Load Test 280
5.19 Presumptive Bearing Capacity 282
5.20 Tolerable Settlement of Buildings 283
Problems 285
References 288
6
Mat Foundations 291
6.1 Introduction 291
6.2 Combined Footings 291
6.3 Common Types of Mat Foundations 294
6.4 Bearing Capacity of Mat Foundations 296
6.5 Differential Settlement of Mats 299
6.6 Field Settlement Observations for Mat Foundations 300
6.7 Compensated Foundation 300
6.8 Structural Design of Mat Foundations 304
Problems 322
References 323
Contents
7
Lateral Earth Pressure 324
7.1 Introduction 324
7.2 Lateral Earth Pressure at Rest 325
Active Pressure
7.3
7.4
7.5
7.6
7.7
7.8
7.9
328
Rankine Active Earth Pressure 328
A Generalized Case for Rankine Active Pressure 334
Coulomb’s Active Earth Pressure 340
Active Earth Pressure Due to Surcharge 348
Active Earth Pressure for Earthquake Conditions 350
Active Pressure for Wall Rotation about the Top: Braced Cut 355
Active Earth Pressure for Translation of Retaining
Wall—Granular Backfill 357
Passive Pressure
360
7.10 Rankine Passive Earth Pressure 360
7.11 Rankine Passive Earth Pressure: Vertical Backface
and Inclined Backfill 363
7.12 Coulomb’s Passive Earth Pressure 365
7.13 Comments on the Failure Surface Assumption for Coulomb’s
Pressure Calculations 366
7.14 Passive Pressure under Earthquake Conditions 370
Problems 371
References 373
8
Retaining Walls 375
8.1 Introduction 375
Gravity and Cantilever Walls
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
377
Proportioning Retaining Walls 377
Application of Lateral Earth Pressure Theories to Design 378
Stability of Retaining Walls 380
Check for Overturning 382
Check for Sliding along the Base 384
Check for Bearing Capacity Failure 387
Construction Joints and Drainage from Backfill 396
Gravity Retaining-Wall Design for Earthquake Conditions 399
Comments on Design of Retaining Walls and a Case Study 402
Mechanically Stabilized Retaining Walls
8.11 Soil Reinforcement 405
405
xi
xii Contents
8.12
8.13
8.14
8.15
Considerations in Soil Reinforcement 406
General Design Considerations 409
Retaining Walls with Metallic Strip Reinforcement 410
Step-by-Step-Design Procedure Using Metallic Strip
Reinforcement 417
8.16 Retaining Walls with Geotextile Reinforcement 422
8.17 Retaining Walls with Geogrid Reinforcement—General 428
8.18 Design Procedure fore Geogrid-Reinforced
Retaining Wall 428
Problems 433
References 435
9
Sheet Pile Walls 437
9.1
9.2
9.3
9.4
9.5
Introduction 437
Construction Methods 441
Cantilever Sheet Pile Walls 442
Cantilever Sheet Piling Penetrating Sandy Soils 442
Special Cases for Cantilever Walls Penetrating
a Sandy Soil 449
9.6 Cantilever Sheet Piling Penetrating Clay 452
9.7 Special Cases for Cantilever Walls Penetrating Clay 457
9.8 Anchored Sheet-Pile Walls 460
9.9 Free Earth Support Method for Penetration
of Sandy Soil 461
9.10 Design Charts for Free Earth Support Method (Penetration into
Sandy Soil) 465
9.11 Moment Reduction for Anchored Sheet-Pile Walls 469
9.12 Computational Pressure Diagram Method for Penetration into
Sandy Soil 472
9.13 Fixed Earth-Support Method for Penetration
into Sandy Soil 476
9.14 Field Observations for Anchor Sheet Pile Walls 479
9.15 Free Earth Support Method for Penetration of Clay 482
9.16 Anchors 486
9.17 Holding Capacity of Anchor Plates in Sand 488
9.18 Holding Capacity of Anchor Plates in Clay
(f 5 0 Condition) 495
9.19 Ultimate Resistance of Tiebacks 495
Problems 497
References 500
Contents
10
Braced Cuts 501
10.1 Introduction 501
10.2 Pressure Envelope for Braced-Cut Design 502
10.3 Pressure Envelope for Cuts in Layered Soil 506
10.4 Design of Various Components of a Braced Cut 507
10.5 Case Studies of Braced Cuts 515
10.6 Bottom Heave of a Cut in Clay 520
10.7 Stability of the Bottom of a Cut in Sand 524
10.8 Lateral Yielding of Sheet Piles and Ground Settlement 529
Problems 531
References 533
11
Pile Foundations 535
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
11.13
11.14
11.15
11.16
11.17
11.18
11.19
Introduction 535
Types of Piles and Their Structural Characteristics 537
Estimating Pile Length 546
Installation of Piles 548
Load Transfer Mechanism 551
Equations for Estimating Pile Capacity 554
Meyerhof’s Method for Estimating Qp 557
Vesic’s Method for Estimating Qp 560
Coyle and Castello’s Method for Estimating Qp in Sand 563
Correlations for Calculating Qp with SPT
and CPT Results 567
Frictional Resistance (Qs) in Sand 568
Frictional (Skin) Resistance in Clay 575
Point Bearing Capacity of Piles Resting on Rock 579
Pile Load Tests 583
Elastic Settlement of Piles 588
Laterally Loaded Piles 591
Pile-Driving Formulas 606
Pile Capacity For Vibration-Driven Piles 611
Negative Skin Friction 613
Group Piles
617
11.20 Group Efficiency 617
11.21 Ultimate Capacity of Group Piles
in Saturated Clay 621
11.22 Elastic Settlement of Group Piles 624
11.23 Consolidation Settlement of Group Piles 626
xiii
xiv
Contents
11.24 Piles in Rock 629
Problems 629
References 634
12
Drilled-Shaft Foundations 637
12.1
12.2
12.3
12.4
12.5
12.6
12.7
Introduction 637
Types of Drilled Shafts 638
Construction Procedures 639
Other Design Considerations 645
Load Transfer Mechanism 646
Estimation of Load-Bearing Capacity 646
Drilled Shafts in Granular Soil: Load-Bearing
Capacity 648
12.8 Load-Bearing Capacity Based on Settlement 652
12.9 Drilled Shafts in Clay: Load-Bearing Capacity 661
12.10 Load-Bearing Capacity Based on Settlement 663
12.11 Settlement of Drilled Shafts at Working Load 668
12.12 Lateral Load-Carrying Capacity—Characteristic Load
and Moment Method 670
12.13 Drilled Shafts Extending into Rock 679
Problems 681
References 685
13
Foundations on Difficult Soils 686
13.1 Introduction 686
Collapsible Soil
686
13.2
13.3
13.4
13.5
Definition and Types of Collapsible Soil 686
Physical Parameters for Identification 687
Procedure for Calculating Collapse Settlement 691
Foundation Design in Soils Not Susceptible
to Wetting 692
13.6 Foundation Design in Soils Susceptible to Wetting 694
Expansive Soils
13.7
13.8
13.9
13.10
695
General Nature of Expansive Soils 695
Unrestrained Swell Test 699
Swelling Pressure Test 700
Classification of Expansive Soil on the Basis
of Index Tests 705
Contents
13.11 Foundation Considerations for Expansive Soils 708
13.12 Construction on Expansive Soils 711
Sanitary Landfills
716
13.13 General Nature of Sanitary Landfills 716
13.14 Settlement of Sanitary Landfills 717
Problems 719
References 720
14
Soil Improvement and Ground Modification 722
14.1 Introduction 722
14.2 General Principles of Compaction 723
14.3 Field Compaction 727
14.4 Compaction Control for Clay Hydraulic Barriers 730
14.5 Vibroflotation 732
14.6 Blasting 739
14.7 Precompression 739
14.8 Sand Drains 745
14.9 Prefabricated Vertical Drains 756
14.10 Lime Stabilization 760
14.11 Cement Stabilization 764
14.12 Fly-Ash Stabilization 766
14.13 Stone Columns 767
14.14 Sand Compaction Piles 772
14.15 Dynamic Compaction 774
14.16 Jet Grouting 776
Problems 778
References 781
Answers to Selected Problems 783
Index 789
xv
Preface
Soil mechanics and foundation engineering have developed rapidly during the last fifty
years. Intensive research and observation in the field and the laboratory have refined and
improved the science of foundation design. Originally published in the fall of 1983 with a
1984 copyright, this text on the principles of foundation engineering is now in the seventh
edition. The use of this text throughout the world has increased greatly over the years; it
also has been translated into several languages. New and improved materials that have
been published in various geotechnical engineering journals and conference proceedings
have been incorporated into each edition of the text.
Principles of Foundation Engineering is intended primarily for undergraduate civil
engineering students. The first chapter, on Geotechnical Properties of Soil, reviews the topics covered in the introductory soil mechanics course, which is a prerequisite for the foundation engineering course. The text is composed of fourteen chapters with examples and
problems, and an answer section for selected problems. The chapters are mostly devoted to
the geotechnical aspects of foundation design. Both Systéime International (SI) units and
English units are used in the text.
Because the text introduces the application of fundamental concepts of foundation
analysis and design to civil engineering students, the mathematical derivations are not
always presented; instead, just the final form of the equation is given. A list of references
for further information and study is included at the end of each chapter.
Each chapter contains many example problems that will help students understand
the application of the equations and graphs. For better understanding and visualization
of the ideas and field practices, about thirty new photographs have been added in this
edition.
A number of practice problems also are given at the end of each chapter. Answers to
some of these problems are given at the end of the text.
The following is a brief overview of the changes from the sixth edition.
•
•
•
•
In several parts of the text, the presentation has been thoroughly reorganized for
better understanding.
A number of new case studies have been added to familiarize students with the
deviations from theory to practice.
In Chapter 1 on Geotechnical Properties of Soil, new sections on liquidity index and
activity have been added. The discussions on hydraulic conductivity of clay, relative
density, and the friction angle of granular soils have been expanded.
Expanded treatment of the weathering process of rocks is given in Chapter 2, Natural
Soil Deposits and Subsoil Exploration.
xvii
xviii
Preface
•
•
•
•
•
•
•
•
•
In Chapter 3 (Shallow Foundations: Ultimate Bearing Capacity), a new case study on
bearing capacity failure in soft saturated clay has been added. Also included is the
reduction factor method for estimating the ultimate bearing capacity of eccentrically
loaded strip foundations on granular soil.
Chapter 4, Ultimate Bearing Capacity of Shallow Foundations: Special Cases, has
new sections on the ultimate bearing capacity of weaker soil underlain by a stronger
soil, the seismic bearing capacity of foundations at the edge of a granular slope,
foundations on rocks, and the stress characteristics solution for foundations located
on the top of granular slopes.
Stress distribution due to a point load and uniformly loaded circular and rectangular
areas located on the surface of a Westergaard-type material has been added to
Chapter 5 on Allowable Bearing Capacity and Settlement. Also included in this
chapter is the procedure to estimate foundation settlement based on Pressuremeter
test results.
Lateral earth pressure due to a surcharge on unyielding retaining structures is now
included in Chapter 7 (Lateral Earth Pressure). Also included in this chapter is the
solution for passive earth pressure on a retaining wall with inclined back face and
horizontal granular backfill using the method of triangular slices.
Chapter 8 on Retaining Walls has a new case study. A more detailed discussion is
provided on the design procedure for geogrid-reinforced retaining walls.
Chapter 9 on Sheet Pile Walls has an added section on the holding capacity of plate
anchors based on the stress characteristics solution.
Two case studies have been added to the chapter on Braced Cuts (Chapter 10).
The chapter on Pile Foundations (Chapter 11) has been thoroughly reorganized for
better understanding.
Based on recent publications, new recommendations have been made to estimate the
load-bearing capacity of drilled shafts extending to rock (Chapter 12).
As my colleagues in the geotechnical engineering area well know, foundation analysis and design is not just a matter of using theories, equations and graphs from a textbook.
Soil profiles found in nature are seldom homogeneous, elastic, and isotropic. The educated
judgment needed to properly apply the theories, equations, and graphs to the evaluation of
soils, foundations, and foundation design cannot be overemphasized or completely taught
in the classroom. Field experience must supplement classroom work.
The following individuals were kind enough to share their photographs which have
been included in this new edition.
•
•
•
•
•
•
•
•
•
•
Professor A. S. Wayal, K. J. Somayia Polytechnic, Mumbai, India
Professor Sanjeev Kumar, Southern Illinois University, Carbondale, Illinois
Mr. Paul J. Koszarek, Professional Service Industries, Inc., Waukesha, Wisconsin
Professor Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida
Professor Jean-Louis Briaud, Texas A&M University, College Station, Texas
Dr. Dharma Shakya, Geotechnical Solutions, Inc., Irvine, California
Mr. Jon Ridgeway, Tensar International, Atlanta, Georgia
Professor N. Sivakugan, James Cook University, Townsville, Queensland, Australia
Professor Anand J. Puppala, University of Texas at Arlington, Arlington, Texas
Professor Thomas M. Petry, Missouri University of Science and Technology,
Rolla, Missouri
Preface
xix
Thanks are due to Neill Belk, graduate student at the University of North Carolina at
Charlotte, and Jennifer Nicks, graduate student at Texas A&M University, College Station,
Texas, for their help during the preparation of this revised edition. I am also grateful for
several helpful suggestions of Professor Adel S. Saada of Case Western Reserve University,
Cleveland, Ohio.
Thanks are due to Chris Carson, Executive Director, Global Publishing Program;
and Hilda Gowans, Senior Developmental Editor, Engineering, Cengage Learning; Lauren
Betsos, Marketing Manager; and Rose Kernan of RPK Editorial Services for their interest
and patience during the revision and production of the manuscript.
For the past twenty-seven years, my primary source of inspiration has been the
immeasurable energy of my wife, Janice. I am grateful for her continual help in the
development of the original text and its six subsequent revisions.
Braja M. Das
1
1.1
Geotechnical Properties of Soil
Introduction
The design of foundations of structures such as buildings, bridges, and dams generally
requires a knowledge of such factors as (a) the load that will be transmitted by the superstructure to the foundation system, (b) the requirements of the local building code, (c) the
behavior and stress-related deformability of soils that will support the foundation system,
and (d) the geological conditions of the soil under consideration. To a foundation engineer,
the last two factors are extremely important because they concern soil mechanics.
The geotechnical properties of a soil—such as its grain-size distribution, plasticity,
compressibility, and shear strength—can be assessed by proper laboratory testing. In addition, recently emphasis has been placed on the in situ determination of strength and deformation properties of soil, because this process avoids disturbing samples during field
exploration. However, under certain circumstances, not all of the needed parameters can
be or are determined, because of economic or other reasons. In such cases, the engineer
must make certain assumptions regarding the properties of the soil. To assess the accuracy
of soil parameters—whether they were determined in the laboratory and the field or
whether they were assumed—the engineer must have a good grasp of the basic principles
of soil mechanics. At the same time, he or she must realize that the natural soil deposits on
which foundations are constructed are not homogeneous in most cases. Thus, the engineer
must have a thorough understanding of the geology of the area—that is, the origin and
nature of soil stratification and also the groundwater conditions. Foundation engineering
is a clever combination of soil mechanics, engineering geology, and proper judgment
derived from past experience. To a certain extent, it may be called an art.
When determining which foundation is the most economical, the engineer must consider the superstructure load, the subsoil conditions, and the desired tolerable settlement.
In general, foundations of buildings and bridges may be divided into two major categories:
(1) shallow foundations and (2) deep foundations. Spread footings, wall footings, and mat
foundations are all shallow foundations. In most shallow foundations, the depth of embedment can be equal to or less than three to four times the width of the foundation. Pile and
drilled shaft foundations are deep foundations. They are used when top layers have poor
1
2 Chapter 1: Geotechnical Properties of Soil
load-bearing capacity and when the use of shallow foundations will cause considerable
structural damage or instability. The problems relating to shallow foundations and mat
foundations are considered in Chapters 3, 4, 5, and 6. Chapter 11 discusses pile foundations, and Chapter 12 examines drilled shafts.
This chapter serves primarily as a review of the basic geotechnical properties of soils.
It includes topics such as grain-size distribution, plasticity, soil classification, effective stress,
consolidation, and shear strength parameters. It is based on the assumption that you have
already been exposed to these concepts in a basic soil mechanics course.
1.2
Grain-Size Distribution
In any soil mass, the sizes of the grains vary greatly. To classify a soil properly, you must
know its grain-size distribution. The grain-size distribution of coarse-grained soil is generally determined by means of sieve analysis. For a fine-grained soil, the grain-size distribution can be obtained by means of hydrometer analysis. The fundamental features of
these analyses are presented in this section. For detailed descriptions, see any soil mechanics laboratory manual (e.g., Das, 2009).
Sieve Analysis
A sieve analysis is conducted by taking a measured amount of dry, well-pulverized soil and
passing it through a stack of progressively finer sieves with a pan at the bottom. The
amount of soil retained on each sieve is measured, and the cumulative percentage of soil
passing through each is determined. This percentage is generally referred to as percent
finer. Table 1.1 contains a list of U.S. sieve numbers and the corresponding size of their
openings. These sieves are commonly used for the analysis of soil for classification
purposes.
Table 1.1 U.S. Standard Sieve Sizes
Sieve No.
Opening (mm)
4
6
8
10
16
20
30
40
50
60
80
100
140
170
200
270
4.750
3.350
2.360
2.000
1.180
0.850
0.600
0.425
0.300
0.250
0.180
0.150
0.106
0.088
0.075
0.053
1.2 Grain-Size Distribution
3
Percent finer (by weight)
100
80
60
40
20
0
10
1
0.1
Grain size, D (mm)
0.01
Figure 1.1 Grain-size distribution
curve of a coarse-grained soil
obtained from sieve analysis
The percent finer for each sieve, determined by a sieve analysis, is plotted on semilogarithmic graph paper, as shown in Figure 1.1. Note that the grain diameter, D, is plotted on
the logarithmic scale and the percent finer is plotted on the arithmetic scale.
Two parameters can be determined from the grain-size distribution curves of coarsegrained soils: (1) the uniformity coefficient (Cu ) and (2) the coefficient of gradation, or
coefficient of curvature (Cc ). These coefficients are
D60
D10
(1.1)
D230
(D60 ) (D10 )
(1.2)
Cu 5
and
Cc 5
where D10, D30, and D60 are the diameters corresponding to percents finer than 10, 30, and
60%, respectively.
For the grain-size distribution curve shown in Figure 1.1, D10 5 0.08 mm,
D30 5 0.17 mm, and D60 5 0.57 mm. Thus, the values of Cu and Cc are
Cu 5
0.57
5 7.13
0.08
and
Cc 5
0.172
5 0.63
(0.57) (0.08)
4 Chapter 1: Geotechnical Properties of Soil
Parameters Cu and Cc are used in the Unified Soil Classification System, which is described
later in the chapter.
Hydrometer Analysis
Hydrometer analysis is based on the principle of sedimentation of soil particles in water.
This test involves the use of 50 grams of dry, pulverized soil. A deflocculating agent is
always added to the soil. The most common deflocculating agent used for hydrometer
analysis is 125 cc of 4% solution of sodium hexametaphosphate. The soil is allowed to
soak for at least 16 hours in the deflocculating agent. After the soaking period, distilled
water is added, and the soil–deflocculating agent mixture is thoroughly agitated. The sample is then transferred to a 1000-ml glass cylinder. More distilled water is added to the
cylinder to fill it to the 1000-ml mark, and then the mixture is again thoroughly agitated.
A hydrometer is placed in the cylinder to measure the specific gravity of the soil–water
suspension in the vicinity of the instrument’s bulb (Figure 1.2), usually over a 24-hour
period. Hydrometers are calibrated to show the amount of soil that is still in suspension at
any given time t. The largest diameter of the soil particles still in suspension at time t can
be determined by Stokes’ law,
D5
18h
L
Å (Gs 2 1)gw Å t
where
D 5 diameter of the soil particle
Gs 5 specific gravity of soil solids
h 5 viscosity of water
L
Figure 1.2 Hydrometer analysis
(1.3)
1.4 Weight–Volume Relationships
5
gw 5 unit weight of water
L 5 effective length (i.e., length measured from the water surface in the cylinder to the
center of gravity of the hydrometer; see Figure 1.2)
t 5 time
Soil particles having diameters larger than those calculated by Eq. (1.3) would have settled
beyond the zone of measurement. In this manner, with hydrometer readings taken at various
times, the soil percent finer than a given diameter D can be calculated and a grain-size distribution plot prepared. The sieve and hydrometer techniques may be combined for a soil
having both coarse-grained and fine-grained soil constituents.
1.3
Size Limits for Soils
Several organizations have attempted to develop the size limits for gravel, sand, silt, and
clay on the basis of the grain sizes present in soils. Table 1.2 presents the size limits recommended by the American Association of State Highway and Transportation Officials
(AASHTO) and the Unified Soil Classification systems (Corps of Engineers, Department
of the Army, and Bureau of Reclamation). The table shows that soil particles smaller than
0.002 mm have been classified as clay. However, clays by nature are cohesive and can be
rolled into a thread when moist. This property is caused by the presence of clay minerals
such as kaolinite, illite, and montmorillonite. In contrast, some minerals, such as quartz
and feldspar, may be present in a soil in particle sizes as small as clay minerals, but these
particles will not have the cohesive property of clay minerals. Hence, they are called claysize particles, not clay particles.
1.4
Weight–Volume Relationships
In nature, soils are three-phase systems consisting of solid soil particles, water, and air (or
gas). To develop the weight–volume relationships for a soil, the three phases can be separated as shown in Figure 1.3a. Based on this separation, the volume relationships can then
be defined.
The void ratio, e, is the ratio of the volume of voids to the volume of soil solids in a
given soil mass, or
Vv
e5
(1.4)
Vs
Table 1.2 Soil-Separate Size Limits
Classification system
Grain size (mm)
Unified
Gravel: 75 mm to 4.75 mm
Sand: 4.75 mm to 0.075 mm
Silt and clay (fines): ,0.075 mm
AASHTO
Gravel: 75 mm to 2 mm
Sand: 2 mm to 0.05 mm
Silt: 0.05 mm to 0.002 mm
Clay: ,0.002 mm
6 Chapter 1: Geotechnical Properties of Soil
Volume
Note: Va + Vw + Vs = V
Ww + Ws = W
Volume
Weight
Air
Va
V
Weight
Vw
Wa = 0
Ww
W
V
Solid
Vs
Ws
(a)
Weight
Volume
Va
Air
V = e
Note: Vw = wGs = Se
Ww = wGsw
Vw = wGs
Vs = 1
Wa = 0
Solid
Ws = Gsw
(b) Unsaturated soil; Vs = 1
Weight
Volume
V = e
Ww = wGsw = ew
Vw = wGs = e
Vs = 1
Solid
Ws = Gsw
(c) Saturated soil; Vs = 1
Figure 1.3 Weight–volume relationships
where
Vv 5 volume of voids
Vs 5 volume of soil solids
The porosity, n, is the ratio of the volume of voids to the volume of the soil specimen, or
n5
where
V 5 total volume of soil
Vv
V
(1.5)
1.4 Weight–Volume Relationships
7
Moreover,
n5
Vv
Vs
Vv
Vv
e
5
5
5
V
V
V
Vs 1 Vv
11e
s
v
1
Vs
Vs
(1.6)
The degree of saturation, S, is the ratio of the volume of water in the void spaces to
the volume of voids, generally expressed as a percentage, or
S(%) 5
Vw
3 100
Vv
(1.7)
where
Vw 5 volume of water
Note that, for saturated soils, the degree of saturation is 100%.
The weight relationships are moisture content, moist unit weight, dry unit weight,
and saturated unit weight, often defined as follows:
Moisture content 5 w(%) 5
Ww
3 100
Ws
(1.8)
where
Ws 5 weight of the soil solids
Ww 5 weight of water
Moist unit weight 5 g 5
W
V
(1.9)
where
W 5 total weight of the soil specimen 5 Ws 1 Ww
The weight of air, Wa, in the soil mass is assumed to be negligible.
Dry unit weight 5 gd 5
Ws
V
(1.10)
When a soil mass is completely saturated (i.e., all the void volume is occupied by
water), the moist unit weight of a soil [Eq. (1.9)] becomes equal to the saturated unit
weight (gsat ). So g 5 gsat if Vv 5 Vw.
More useful relations can now be developed by considering a representative soil specimen in which the volume of soil solids is equal to unity, as shown in Figure 1.3b. Note that
if Vs 5 1, then, from Eq. (1.4), Vv 5 e, and the weight of the soil solids is
Ws 5 Gsgw
where
Gs 5 specific gravity of soil solids
gw 5 unit weight of water (9.81 kN>m3)
8
Chapter 1: Geotechnical Properties of Soil
Also, from Eq. (1.8), the weight of water Ww 5 wWs. Thus, for the soil specimen under
consideration, Ww 5 wWs 5 wGsgw. Now, for the general relation for moist unit weight
given in Eq. (1.9),
Ws 1 Ww
Gsgw (1 1 w)
W
5
5
V
Vs 1 Vv
11e
g5
(1.11)
Similarly, the dry unit weight [Eq. (1.10)] is
gd 5
Ws
Ws
Gsgw
5
5
V
Vs 1 Vv
11e
(1.12)
From Eqs. (1.11) and (1.12), note that
gd 5
g
11w
(1.13)
According to Eq. (1.7), degree of saturation is
S5
Vw
Vv
Now, referring to Fig. 1.3(b),
Vw 5 wGs
and
Vv 5 e
Thus,
S5
Vw
wGs
5
e
Vv
(1.14)
For a saturated soil, S ⫽ 1. So
e 5 wGs
(1.15)
The saturated unit weight of soil then becomes
gsat 5
Ws 1 Ww
Gsgw 1 e gw
5
Vs 1 Vv
11e
(1.16)
1.4 Weight–Volume Relationships
9
In SI units, Newton or kiloNewton is weight and is a derived unit, and g or kg is
mass. The relationships given in Eqs. (1.11), (1.12) and (1.16) can be expressed as moist,
dry, and saturated densities as follow:
r5
Gsrw (1 1 w)
11e
(1.17)
Gsrw
11e
(1.18)
rd 5
r sat 5
rw (Gs 1 e)
11e
(1.19)
where ␳, ␳d, ␳sat ⫽ moist density, dry density, and saturated density, respectively
␳w ⫽ density of water (⫽ 1000 kg/m3)
Relationships similar to Eqs. (1.11), (1.12), and (1.16) in terms of porosity can also
be obtained by considering a representative soil specimen with a unit volume (Figure 1.3c).
These relationships are
g 5 Gsgw (1 2 n) (1 1 w)
(1.20)
gd 5 (1 2 n)Gsgw
(1.21)
gsat 5 3(1 2 n)Gs 1 n4gw
(1.22)
and
Table 1.3 gives a summary of various forms of relationships that can be obtained for
˝, ˝d, and ˝sat.
Table 1.3 Various Forms of Relationships for ␥, ␥d, and ␥sat
Unit-weight relationship
Dry unit weight
(1 1 w)Gsgw
11e
(Gs 1 Se)gw
g5
11e
(1 1 w)Gsgw
g5
wGs
11
S
␥ ⫽ Gs␥w(1 ⫺ n)(1 + w)
g
11w
Gsgw
gd 5
11e
␥d ⫽ Gs␥w(1 ⫺ n)
Gs
g
gd 5
wGs w
11
S
eSgw
gd 5
(1 1 e)w
g5
gd 5
␥d ⫽ ␥sat ⫺ n␥w
gd 5 gsat 2 a
e
bg
11e w
Saturated unit weight
(Gs 1 e)gw
11e
␥sat ⫽ [(1 ⫺ n)Gs + n]␥w
gsat 5
gsat 5 a
11w
bG g
1 1 wGs s w
e
11w
bg
gsat 5 a b a
w
11e w
␥sat ⫽ ␥d + n␥w
gsat 5 gd 1 a
e
bg
11e w
10 Chapter 1: Geotechnical Properties of Soil
Table 1.4 Specific Gravities of Some Soils
Type of soil
Gs
Quartz sand
Silt
Clay
Chalk
Loess
Peat
2.64–2.66
2.67–2.73
2.70–2.9
2.60–2.75
2.65–2.73
1.30–1.9
Except for peat and highly organic soils, the general range of the values of specific
gravity of soil solids (Gs ) found in nature is rather small. Table 1.4 gives some representative values. For practical purposes, a reasonable value can be assumed in lieu of running
a test.
1.5
Relative Density
In granular soils, the degree of compaction in the field can be measured according to the
relative density, defined as
Dr (%) 5
emax 2 e
3 100
emax 2 emin
(1.23)
where
emax 5 void ratio of the soil in the loosest state
emin 5 void ratio in the densest state
e 5 in situ void ratio
The relative density can also be expressed in terms of dry unit weight, or
Dr (%) 5 b
gd 2 gd(min)
gd(max) 2 gd(min)
r
gd(max)
gd
3 100
(1.24)
where
gd 5 in situ dry unit weight
gd(max) 5 dry unit weight in the densest state; that is, when the void ratio is emin
gd(min) 5 dry unit weight in the loosest state; that is, when the void ratio is emax
The denseness of a granular soil is sometimes related to the soil’s relative density.
Table 1.5 gives a general correlation of the denseness and Dr. For naturally occurring
sands, the magnitudes of emax and emin [Eq. (1.23)] may vary widely. The main reasons for
such wide variations are the uniformity coefficient, Cu, and the roundness of the particles.
1.5 Relative Density
11
Table 1.5 Denseness of a Granular Soil
Relative density, Dr (%)
Description
0–20
20–40
40–60
60–80
80–100
Very loose
Loose
Medium
Dense
Very dense
Example 1.1
The moist weight of 28.3 ⫻ 10⫺4 m3 of soil is 54.27 N. If the moisture content is 12%
and the specific gravity of soil solids is 2.72, find the following:
a.
b.
c.
d.
e.
f.
Moist unit weight (kN/m3)
Dry unit weight (kN/m3)
Void ratio
Porosity
Degree of saturation (%)
Volume occupied by water (m3)
Solution
Part a
From Eq. (1.9),
g5
W
54.27 3 1023
5
5 19.18 kN>m3
V
0.00283
Part b
From Eq. (1.13),
gd 5
g
19.18
5
5 17.13 kN>m3
11w
12
11
100
Part c
From Eq. (1.12),
gd 5
or
17.13 5
Gsgw
11e
(2.72) (9.81)
11e
e 5 0.56
12 Chapter 1: Geotechnical Properties of Soil
Part d
From Eq. (1.6),
n5
e
0.56
5
5 0.359
11e
1 1 0.56
Part e
From Eq. (1.14),
S5
wGs
(0.12) (2.72)
5 0.583
5
e
0.56
Part f
From Eq. (1.12),
Ws 5
W
54.27 3 1023
5
5 48.46 3 1023 kN
11w
1.12
Ww 5 W 2 Ws 5 54.27 3 1023 2 48.46 3 1023 5 5.81 3 1023 kN
Vw 5
5.81 3 1023
5 0.592 3 1023 m3
9.81
■
Example 1.2
The dry density of a sand with a porosity of 0.387 is 1600 kg/m3. Find the void ratio of
the soil and the specific gravity of the soil solids.
Solution
Void ratio
Given: n ⫽ 0.387. From Eq. (1.6),
n
0.387
e5
5
5 0.631
12n
1 2 0.387
Specific gravity of soil solids
From Eq. (1.18),
rd 5
1600 5
Gsrw
11e
Gs (1000)
1.631
Gs 5 2.61
■
1.5 Relative Density
13
Example 1.3
The moist unit weight of a soil is 19.2 kN/m3. Given Gs ⫽ 2.69 and moisture content
w 5 9.8%, determine
a.
b.
c.
d.
Dry unit weight (kN/m3)
Void ratio
Porosity
Degree of saturation (%)
Solution
Part a
From Eq. (1.13),
gd 5
g
5
11w
19.2
5 17.49 kN>m3
9.8
11
100
Part b
From Eq. (1.12),
gd 5 17.49 kN>m3 5
Gsgw
(2.69) (9.81)
5
11e
11e
e 5 0.509
Part c
From Eq. (1.6),
n5
e
0.509
5
5 0.337
11e
1 1 0.509
Part d
From Eq. (1.14),
S5
(0.098) (2.69)
wGs
5 c
d (100) 5 51.79%
e
0.509
■
Example 1.4
For a saturated soil, given w ⫽ 40% and Gs ⫽ 2.71, determine the saturated and dry
unit weights in lb/ft3 and kN/m3.
Solution
For saturated soil, from Eq. (1.15),
e ⫽ wGs ⫽ (0.4)(2.71) ⫽ 1.084
14 Chapter 1: Geotechnical Properties of Soil
From Eq. (1.16),
gsat 5
(Gs 1 e)gw
(2.71 1 1.084)9.81
5
5 17.86 kN>m3
11e
1 1 1.084
From Eq. (1.12),
gd 5
Gsgw
(2.71) (9.81)
5
5 12.76 kN>m3
11e
1 1 1.084
■
Example 1.5
The mass of a moist soil sample collected from the field is 465 grams, and its oven dry
mass is 405.76 grams. The specific gravity of the soil solids was determined in the
laboratory to be 2.68. If the void ratio of the soil in the natural state is 0.83, find the
following:
a. The moist density of the soil in the field (kg/m3)
b. The dry density of the soil in the field (kg/m3)
c. The mass of water, in kilograms, to be added per cubic meter of soil in the field
for saturation
Solution
Part a
From Eq. (1.8),
w5
Ww
465 2 405.76
59.24
5
5
5 14.6%
Ws
405.76
405.76
From Eq. (1.17),
r5
Gsrw 1 wGsrw
Gsrw (1 1 w)
(2.68) (1000) (1.146)
5
5
11e
11e
1.83
5 1678.3 kg>m3
Part b
From Eq. (1.18),
rd 5
Gsrw
(2.68) (1000)
5
5 1468.48 kg>m3
11e
1.83
1.6 Atterberg Limits
15
Part c
Mass of water to be added ⫽ ␳sat ⫺ ␳
From Eq. (1.19),
rsat 5
Gsrw 1 erw
rw (Gs 1 e)
(1000) (2.68 1 0.83)
5
5
5 1918 kg>m3
11e
11e
1.83
So, mass of water to be added ⫽ 1918 ⫺ 1678.3 ⫽ 239.7 kg/m3.
■
Example 1.6
The maximum and minimum dry unit weights of a sand are 17.1 kN>m3 and
14.2 kN>m3, respectively. The sand in the field has a relative density of 70% with a
moisture content of 8%. Determine the moist unit weight of the sand in the field.
Solution
From Eq. (1.24),
gd 2 gd(min)
dc
gd(max)
Dr 5 c
gd(max) 2 gd(min)
0.7 5 c
gd 2 14.2
17.1
dc
d
gd
17.1 2 14.2
gd
d
gd 5 16.11 kN>m3
g 5 gd (1 1 w) 5 16.11a1 1
1.6
8
b 5 17.4 kN , m3
100
■
Atterberg Limits
When a clayey soil is mixed with an excessive amount of water, it may flow like a semiliquid.
If the soil is gradually dried, it will behave like a plastic, semisolid, or solid material, depending on its moisture content. The moisture content, in percent, at which the soil changes from
a liquid to a plastic state is defined as the liquid limit (LL). Similarly, the moisture content,
in percent, at which the soil changes from a plastic to a semisolid state and from a semisolid
to a solid state are defined as the plastic limit (PL) and the shrinkage limit (SL), respectively.
These limits are referred to as Atterberg limits (Figure 1.4):
•
•
The liquid limit of a soil is determined by Casagrande’s liquid device (ASTM Test
Designation D-4318) and is defined as the moisture content at which a groove
closure of 12.7 mm occurs at 25 blows.
The plastic limit is defined as the moisture content at which the soil crumbles when
rolled into a thread of 3.18 mm in diameter (ASTM Test Designation D-4318).
16 Chapter 1: Geotechnical Properties of Soil
Solid
state
Semisolid
state
Plastic
state
Semiliquid
state Increase of
moisture content
Volume of the
soil–water
mixture
SL
PL
LL
Moisture
content
Figure 1.4 Definition of Atterberg limits
•
The shrinkage limit is defined as the moisture content at which the soil does not
undergo any further change in volume with loss of moisture (ASTM Test
Designation D-427).
The difference between the liquid limit and the plastic limit of a soil is defined as the
plasticity index (PI), or
PI 5 LL 2 PL
1.7
(1.25)
Liquidity Index
The relative consistency of a cohesive soil in the natural state can be defined by a ratio
called the liquidity index, which is given by
LI 5
w 2 PL
LL 2 PL
(1.26)
where w ⫽ in situ moisture content of soil.
The in situ moisture content for a sensitive clay may be greater than the liquid limit.
In this case,
LI ⬎ 1
These soils, when remolded, can be transformed into a viscous form to flow like a
liquid.
1.9 Soil Classification Systems
17
Soil deposits that are heavily overconsolidated may have a natural moisture content
less than the plastic limit. In this case,
LI ⬍ 0
1.8
Activity
Because the plasticity of soil is caused by the adsorbed water that surrounds the clay particles, we can expect that the type of clay minerals and their proportional amounts in a soil
will affect the liquid and plastic limits. Skempton (1953) observed that the plasticity index
of a soil increases linearly with the percentage of clay-size fraction (% finer than 2 ␮m by
weight) present. The correlations of PI with the clay-size fractions for different clays plot
separate lines. This difference is due to the diverse plasticity characteristics of the various
types of clay minerals. On the basis of these results, Skempton defined a quantity called
activity, which is the slope of the line correlating PI and % finer than 2 ␮m. This activity
may be expressed as
A5
PI
(% of clay-size fraction, by weight)
(1.27)
Activity is used as an index for identifying the swelling potential of clay soils.
Typical values of activities for various clay minerals are given in Table 1.6.
1.9
Soil Classification Systems
Soil classification systems divide soils into groups and subgroups based on common engineering properties such as the grain-size distribution, liquid limit, and plastic limit. The
two major classification systems presently in use are (1) the American Association of State
Highway and Transportation Officials (AASHTO) System and (2) the Unified Soil
Classification System (also ASTM). The AASHTO system is used mainly for the classification of highway subgrades. It is not used in foundation construction.
Table 1.6 Activities of Clay Minerals
Mineral
Smectites
Illite
Kaolinite
Halloysite (4H2O)
Halloysite (2H2O)
Attapulgite
Allophane
Activity (A)
1–7
0.5–1
0.5
0.5
0.1
0.5–1.2
0.5–1.2
18 Chapter 1: Geotechnical Properties of Soil
AASHTO System
The AASHTO Soil Classification System was originally proposed by the Highway Research
Board’s Committee on Classification of Materials for Subgrades and Granular Type Roads
(1945). According to the present form of this system, soils can be classified according to eight
major groups, A-1 through A-8, based on their grain-size distribution, liquid limit, and plasticity indices. Soils listed in groups A-1, A-2, and A-3 are coarse-grained materials, and those
in groups A-4, A-5, A-6, and A-7 are fine-grained materials. Peat, muck, and other highly
organic soils are classified under A-8. They are identified by visual inspection.
The AASHTO classification system (for soils A-1 through A-7) is presented in
Table 1.7. Note that group A-7 includes two types of soil. For the A-7-5 type, the plasticity
Table 1.7 AASHTO Soil Classification System
Granular materials
(35% or less of total sample passing No. 200 sieve)
General classification
A-1
Group classification
Sieve analysis (% passing)
No. 10 sieve
No. 40 sieve
No. 200 sieve
For fraction passing
No. 40 sieve
Liquid limit (LL)
Plasticity index (PI)
Usual type of material
A-2
A-1-a
A-1-b
A-3
A-2-4
A-2-5
A-2-6
A-2-7
50 max
30 max
15 max
50 max
25 max
51 min
10 max
35 max
35 max
35 max
35 max
Nonplastic
Fine sand
40 max
41 min
40 max
41 min
10 max
10 max
11 min
11 min
Silty or clayey gravel and sand
6 max
Stone fragments,
gravel, and sand
Subgrade rating
Excellent to good
Silt–clay materials
(More than 35% of total sample passing No. 200 sieve)
General classification
Group classification
A-4
A-5
A-6
A-7
A-7-5a
A-7-6b
Sieve analysis (% passing)
No. 10 sieve
No. 40 sieve
No. 200 sieve
For fraction passing
No. 40 sieve
Liquid limit (LL)
Plasticity index (PI)
Usual types of material
Subgrade rating
a
If PI < LL 2 30, the classification is A-7-5.
If PI . LL 2 30, the classification is A-7-6.
b
36 min
36 min
36 min
40 max
41 min
10 max
10 max
Mostly silty soils
36 min
40 max
41 min
11 min
11 min
Mostly clayey soils
Fair to poor
1.9 Soil Classification Systems
19
index of the soil is less than or equal to the liquid limit minus 30. For the A-7-6 type, the
plasticity index is greater than the liquid limit minus 30.
For qualitative evaluation of the desirability of a soil as a highway subgrade material, a number referred to as the group index has also been developed. The higher the value
of the group index for a given soil, the weaker will be the soil’s performance as a subgrade.
A group index of 20 or more indicates a very poor subgrade material. The formula for the
group index is
GI 5 (F200 2 35) 30.2 1 0.005(LL 2 40) 4 1 0.01(F200 2 15) (PI 2 10)
(1.28)
where
F200 5 percent passing No. 200 sieve, expressed as a whole number
LL 5 liquid limit
PI 5 plasticity index
When calculating the group index for a soil belonging to group A-2-6 or A-2-7, use only
the partial group-index equation relating to the plasticity index:
GI 5 0.01(F200 2 15) (PI 2 10)
(1.29)
The group index is rounded to the nearest whole number and written next to the soil group
in parentheses; for example, we have
A-4
()*
Z
Soil group
(5)
()*
Group index
The group index for soils which fall in groups A-1-a, A-1-b, A-3, A-2-4, and A-2-5 is
always zero.
Unified System
The Unified Soil Classification System was originally proposed by A. Casagrande in
1942 and was later revised and adopted by the United States Bureau of Reclamation
and the U.S. Army Corps of Engineers. The system is currently used in practically all
geotechnical work.
In the Unified System, the following symbols are used for identification:
Symbol
G
S
M
Description
Gravel Sand Silt
C
O
Pt
H
L
W
Clay
Organic silts
and clay
Peat and highly
organic soils
High
Low
Well
plasticity plasticity graded
P
Poorly
graded
20 Chapter 1: Geotechnical Properties of Soil
70
Plasticity index, PI
60
U-line
PI 0.9 (LL 8)
50
40
CL
or
OL
30
20
CL ML
ML
or
OL
10
0
0
10
20
30
CH
or
OH
A-line
PI 0.73 (LL 20)
MH
or
OH
40 50 60 70
Liquid limit, LL
80
90
100
Figure 1.5 Plasticity chart
The plasticity chart (Figure 1.5) and Table 1.8 show the procedure for determining the
group symbols for various types of soil. When classifying a soil be sure to provide the group
name that generally describes the soil, along with the group symbol. Figures 1.6, 1.7, and
1.8 give flowcharts for obtaining the group names for coarse-grained soil, inorganic finegrained soil, and organic fine-grained soil, respectively.
Example 1.7
Classify the following soil by the AASHTO classification system.
Percent passing No. 4 sieve 5 92
Percent passing No. 10 sieve 5 87
Percent passing No. 40 sieve 5 65
Percent passing No. 200 sieve 5 30
Liquid limit 5 22
Plasticity index 5 8
Solution
Table 1.7 shows that it is a granular material because less than 35% is passing a No. 200
sieve. With LL 5 22 (that is, less than 40) and PI 5 8 (that is, less than 10), the soil falls
in group A-2-4.
The soil is A-2-4(0).
■
21
b
Cu 5 D60>D10
Cc 5
(D30 ) 2
j
, 0.75
PT
OH
Organic silt k, l, m, q
Peat
Organic clay k, l, m, p
Elastic silt k , l , m
Fat clay k , l , m
Organic silt k, l, m, o
Organic clay k, l, m, n
Siltk , l , m
Lean clayk , l , m
Clayey sandg , h , i
Silty sandg , h , i
Poorly graded sandi
Well-graded sandi
Clayey gravel f , g , h
Silty gravel f , g , h
Poorly graded gravel f
Well-graded gravel f
k
If soil contains 15 to 29% plus No. 200, add “with sand”
or “with gravel,” whichever is predominant.
l
If soil contains >30% plus No. 200, predominantly sand,
add “sandy” to group name.
m
If soil contains >30% plus No. 200, predominantly
gravel, add “gravelly” to group name.
n
PI > 4 and plots on or above “A” line.
o
PI , 4 or plots below “A” line.
p
PI plots on or above “A” line.
q
PI plots below “A” line.
Liquid limit—not dried
Liquid limit—oven dried
MH
PI plots below “A” line
OL
CH
, 0.75
ML
CL
PI plots on or above “A” line
Liquid limit—not dried
Liquid limit—oven dried
PI , 4 or plots below “A” line j
PI . 7 and plots on or above “A” line
SC
SP
Cu , 6 and/or 1 . Cc . 3e
Fines classify as CL or CH
SW
SM
GC
Fines classify as CL or CH
Cu > 6 and 1 < Cc < 3e
Fines classify as ML or MH
GM
GP
Cu , 4 and/or 1 . Cc . 3e
Fines classify as ML or MH
GW
Cu > 4 and 1 < Cc < 3e
Primarily organic matter, dark in color, and organic odor
Organic
Inorganic
Organic
Inorganic
Sand with Fines
More than 12% finesd
Clean Sands
Less than 5% finesd
Gravels with Fines
More than 12% finesc
Clean Gravels
Less than 5% finesc
D10 3 D60
f
If soil contains >15% sand, add “with sand” to group
name.
g
If fines classify as CL-ML, use dual symbol GC-GM or
SC-SM.
h
If fines are organic, add “with organic fines” to group
name.
i
If soil contains >15% gravel, add “with gravel” to group
name.
j
If Atterberg limits plot in hatched area, soil is a
CL-ML, silty clay.
e
Silts and Clays
Liquid limit 50 or more
Silts and Clays
Liquid limit less than 50
Sands
50% or more of coarse
fraction passes No. 4 sieve
Gravels
More than 50% of coarse
fraction retained on No. 4
sieve
Based on the material passing the 75-mm. (3-in) sieve.
If field sample contained cobbles or boulders, or both,
add “with cobbles or boulders, or both” to group name.
c
Gravels with 5 to 12% fines require dual symbols: GWGM well-graded gravel with silt; GW-GC well-graded
gravel with clay; GP-GM poorly graded gravel with silt;
GP-GC poorly graded gravel with clay.
d
Sands with 5 to 12% fines require dual symbols: SWSM well-graded sand with silt; SW-SC well-graded sand
with clay; SP-SM poorly graded sand with silt; SP-SC
poorly graded sand with clay.
a
Highly organic soils
Fine-grained soils
50% or more passes the
No. 200 sieve
Coarse-grained soils
More than 50% retained on
No. 200 sieve
Criteria for assigning group symbols and group names using laboratory testsa
Soil classification
Group
symbol
Group nameb
Table 1.8 Unified Soil Classification Chart (after ASTM, 2009) (ASTM D2487-98: Standard Practice for Classification of Soils for Engineering Purposes
(Unified Soil Classification). Copyright ASTM INTERNATIONAL. Reprinted with permission.)
22
12% fines
5–12% fines
5% fines
12% fines
5–12% fines
fines ML or MH
GC
GC-GM
fines CL-ML
Cu 6 and/or 1 Cc 3
SW-SM
SM
SC
SC-SM
fines CL or CH
fines CL-ML
SP-SC
SP-SM
SW-SC
fines ML or MH
fines CL, CH
(or CL-ML)
fines ML or MH
fines CL, CH
(or CL-ML)
SP
Cu 6 and/or 1 Cc 3
Cu 6 and 1 Cc 3
SW
fines ML or MH
GM
fines CL or CH
GP-GC
GP-GM
fines ML or MH
fines CL, CH
(or CL-ML)
fines ML or MH
GW-GM
GW-GC
Cu 6 and 1 Cc 3
Cu 4 and/or 1 Cc 3
fines CL, CH
(or CL-ML)
GP
Cu 4 and/or 1 Cc 3
Cu 4 and 1 Cc 3
GW
Cu 4 and 1 Cc 3
Silty gravel
Silty gravel with sand
Clayey gravel
Clayey gravel with sand
Silty, clayey gravel
Silty, clayey gravel with sand
15% sand
15% sand
15% sand
15% sand
15% sand
15% sand
Well-graded sand with silt
Well-graded sand with silt and gravel
Well-graded sand with clay (or silty clay)
Well-graded sand with clay and gravel (or silty clay and gravel)
Poorly graded sand with silt
Poorly graded sand with silt and gravel
Poorly graded sand with clay (or silty clay)
Poorly graded sand with clay and gravel (or silty clay and gravel)
Silty sand
Silty sand with gravel
Clayey sand
Clayey sand with gravel
Silty, clayey sand
Silty, clayey sand with gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
15% gravel
Well-graded sand
Well-graded sand with gravel
Poorly graded sand
Poorly graded sand with gravel
Poorly graded gravel with silt
Poorly graded gravel with silt and sand
Poorly graded gravel with clay (or silty clay)
Poorly graded gravel with clay and sand (or silty clay and sand)
15% sand
15% sand
15% sand
15% sand
15% gravel
15% gravel
15% gravel
15% gravel
Well-graded gravel with silt
Well-graded gravel with silt and sand
Well-graded gravel with clay (or silty clay)
Well-graded gravel with clay and sand (or silty clay and sand)
Well-graded gravel
Well-graded gravel with sand
Poorly graded gravel
Poorly graded gravel with sand
15% sand
15% sand
15% sand
15% sand
15% sand
15% sand
15% sand
15% sand
Group Name
Figure 1.6 Flowchart for classifying coarse-grained soils (more than 50% retained on No. 200 Sieve) (After ASTM, 2009) (ASTM D2487-98:
Standard Practice for Classification of Soils for Engineering Purposes (Unified Soil Classification). Copyright ASTM INTERNATIONAL.
Reprinted with permission.)
Sand
% sand % gravel
Gravel
% gravel % sand
5% fines
Group Symbol
23
Organic
Inorganic
Organic
(
(
LL—ovendried
0.75
LL—not dried
PI plots below
“A”—line
PI plots on or
above “A”—line
LL—ovendried
0.75
LL—not dried
PI 4 or plots
below “A”—line
)
)
OH
MH
CH
OL
ML
CL-ML
CL
See Figure 1.8
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
See figure 1.8
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15-29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15 −29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
Elastic silt
Elastic silt with sand
Elastic silt with gravel
Sandy elastic silt
Sandy elastic silt with gravel
Gravelly elastic silt
Gravelly elastic silt with sand
Fat clay
Fat clay with sand
Fat clay with gravel
Sandy fat clay
Sandy fat clay with gravel
Gravelly fat clay
Gravelly fat clay with sand
Silt
Silt with clay
Silt with gravel
Sandy silt
Sandy silt with gravel
Gravelly silt
Gravelly silt with sand
Silty clay
Silty clay with sand
Silty clay with gravel
Sandy silty clay
Sandy silty clay with gravel
Gravelly silty clay
Gravelly silty clay with sand
Lean clay
Lean clay with sand
Lean clay with gravel
Sandy lean clay
Sandy lean clay with gravel
Gravelly lean clay
Gravelly lean clay with sand
Group Name
Figure 1.7 Flowchart for classifying fine-grained soil (50% or more passes No. 200 Sieve) (After ASTM, 2009)(ASTM D2487-98:
Standard Practice for Classification of Soils for Engineering Purposes (Unified Soil Classification). Copyright ASTM INTERNATIONAL.
Reprinted with permission.)
LL 50
LL 50
Inorganic
4 PI 7 and
plots on or above
“A”—line
PI 7 and plots
on or above
A
ì ”—line
Group Symbol
24
Plots below
“A”—line
Plots on or
above “A”—line
PI 4 and plots
below “A”—line
PI 4 and plots
on or above “A”—line
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
30% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% plus No. 200
15−29% plus No. 200
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
% sand % gravel
% sand % gravel
15% gravel
15% gravel
15% sand
15% sand
Organic silt
Organic silt with sand
Organic silt with gravel
Sandy organic silt
Sandy organic silt with gravel
Gravelly organic silt
Gravelly organic silt with sand
Organic clay
Organic clay with sand
Organic clay with gravel
Sandy organic clay
Sandy organic clay with gravel
Gravelly organic clay
Gravelly organic clay with sand
Organic silt
Organic silt with sand
Organic silt with gravel
Sandy organic silt
Sandy organic silt with gravel
Gravelly organic silt
Gravelly organic silt with sand
Organic clay
Organic clay with sand
Organic clay with gravel
Sandy organic clay
Sandy organic clay with gravel
Gravelly organic clay
Gravelly organic clay with sand
Group Name
Figure 1.8 Flowchart for classifying organic fine-grained soil (50% or more passes No. 200 Sieve) (After ASTM, 2009)
(ASTM D2487-98: Standard Practice for Classification of Soils for Engineering Purposes (Unified Soil Classification). Copyright ASTM
INTERNATIONAL. Reprinted with permission.)
OH
OL
Group Symbol
F
1.10 Hydraulic Conductivity of Soil
25
Example 1.8
Classify the following soil by the Unified Soil Classification System:
Percent passing No. 4 sieve 5 82
Percent passing No. 10 sieve 5 71
Percent passing No. 40 sieve 5 64
Percent passing No. 200 sieve 5 41
Liquid limit 5 31
Plasticity index 5 12
Solution
We are given that F200 5 41, LL 5 31, and PI 5 12. Since 59% of the sample is retained on a No. 200 sieve, the soil is a coarse-grained material. The percentage passing
a No. 4 sieve is 82, so 18% is retained on No. 4 sieve (gravel fraction). The coarse fraction passing a No. 4 sieve (sand fraction) is 59 2 18 5 41% (which is more than 50%
of the total coarse fraction). Hence, the specimen is a sandy soil.
Now, using Table 1.8 and Figure 1.5, we identify the group symbol of the soil
as SC.
Again from Figure 1.6, since the gravel fraction is greater than 15%, the group
name is clayey sand with gravel.
■
1.10
Hydraulic Conductivity of Soil
The void spaces, or pores, between soil grains allow water to flow through them. In soil
mechanics and foundation engineering, you must know how much water is flowing through
a soil per unit time. This knowledge is required to design earth dams, determine the quantity of seepage under hydraulic structures, and dewater foundations before and during their
construction. Darcy (1856) proposed the following equation (Figure 1.9) for calculating the
velocity of flow of water through a soil:
v 5 ki
(1.30)
In this equation,
v 5 Darcy velocity (unit: cm>sec)
k 5 hydraulic conductivity of soil (unit: cm>sec)
i 5 hydraulic gradient
The hydraulic gradient is defined as
i5
Dh
L
where
Dh 5 piezometric head difference between the sections at AA and BB
L 5 distance between the sections at AA and BB
(Note: Sections AA and BB are perpendicular to the direction of flow.)
(1.31)
26 Chapter 1: Geotechnical Properties of Soil
h
A
B
Direction
of flow
Soil
A
Direction
of flow
L
B
Figure 1.9 Definition of
Darcy’s law
Darcy’s law [Eq. (1.30)] is valid for a wide range of soils. However, with materials
like clean gravel and open-graded rockfills, the law breaks down because of the turbulent
nature of flow through them.
The value of the hydraulic conductivity of soils varies greatly. In the laboratory, it can
be determined by means of constant-head or falling-head permeability tests. The constanthead test is more suitable for granular soils. Table 1.9 provides the general range for the values of k for various soils. In granular soils, the value depends primarily on the void ratio. In the
past, several equations have been proposed to relate the value of k to the void ratio in granular
soil. However the author recommends the following equation for use (also see Carrier, 2003):
e3
k~
(1.32)
11e
where
k 5 hydraulic conductivity
e 5 void ratio
Chapuis (2004) proposed an empirical relationship for k in conjunction with
Eq. (1.32) as
k(cm>s) 5 2.4622 cD210
0.7825
e3
d
(1 1 e)
where D 5 effective size (mm).
Table 1.9 Range of the Hydraulic Conductivity for Various Soils
Type of soil
Medium to coarse gravel
Coarse to fine sand
Fine sand, silty sand
Silt, clayey silt, silty clay
Clays
Hydraulic
conductivity, k
(cm/sec)
Greater than 10 21
10 21 to 10 23
10 23 to 10 25
10 24 to 10 26
10 27 or less
(1.33)
1.10 Hydraulic Conductivity of Soil
27
The preceding equation is valid for natural, uniform sand and gravel to predict k that
is in the range of 1021 to 1023 cm>s. This can be extended to natural, silty sands without
plasticity. It is not valid for crushed materials or silty soils with some plasticity.
Based on laboratory experimental results, Amer and Awad (1974) proposed the following relationship for k in granular soil:
k 5 3.5 3 1024 a
rw
e3
2.32
b
bC 0.6
u D 10 a
h
11e
(1.34)
where
k is in cm/sec
Cu ⫽ uniformity coefficient
D10 ⫽ effective size (mm)
␳w ⫽ density of water (g/cm3)
␩ ⫽ viscosity (g⭈s/cm2)
At 20°C, ␳w ⫽ 1 g/cm3 and ␩ ⬇ 0.1 ⫻ 10⫺4 g⭈s/cm2. So
k 5 3.5 3 1024 a
e3
1
bC 0.6D2.32 a
b
1 1 e u 10 0.1 3 1024
or
k (cm>sec) 5 35a
e3
bC 0.6D2.32
1 1 e u 10
(1.35)
Hydraulic Conductivity of Cohesive Soil
According to their experimental observations, Samarasinghe, Huang, and Drnevich (1982)
suggested that the hydraulic conductivity of normally consolidated clays could be given by
the equation
k5C
en
11e
(1.36)
where C and n are constants to be determined experimentally.
Some other empirical relationships for estimating the hydraulic conductivity in clayey
soils are given in Table 1.10. One should keep in mind, however, that any empirical
Table 1.10 Empirical Relationships for Estimating Hydraulic Conductivity in Clayey Soil
Type of soil
Source
Relationshipa
Clay
Mesri and Olson (1971)
log k ⫽ A⬘ log e + B⬘
e0 2 e
log k 5 log k0 2
Ck
Ck ⬇ 0.5e0
Taylor (1948)
k0 ⫽ in situ hydraulic conductivity at void ratio e0
k ⫽ hydraulic conductivity at void ratio e
Ck ⫽ hydraulic conductivity change index
a
28 Chapter 1: Geotechnical Properties of Soil
PI + CF = 1.25
2.8
1.0
2.4
Void ratiuo, e
2.0
0.75
1.6
1.2
0.5
0.8
0.4
10–11
10–10
10–9
5 × 10–9
k (m/sec)
Figure 1.10 Variation of void ratio with hydraulic conductivity of clayey
soils (Based on Tavenas, et al., 1983)
relationship of this type is for estimation only, because the magnitude of k is a highly variable parameter and depends on several factors.
Tavenas, et al. (1983) also gave a correlation between the void ratio and the
hydraulic conductivity of clayey soil. This correlation is shown in Figure 1.10. An important point to note, however, is that in Figure 6.10, PI, the plasticity index, and CF, the claysize fraction in the soil, are in fraction (decimal) form.
1.11
Steady-State Seepage
For most cases of seepage under hydraulic structures, the flow path changes direction and
is not uniform over the entire area. In such cases, one of the ways of determining the rate
of seepage is by a graphical construction referred to as the flow net, a concept based on
Laplace’s theory of continuity. According to this theory, for a steady flow condition, the
flow at any point A (Figure 1.11) can be represented by the equation
kx
'2h
'2h
'2h
1
k
1
k
50
y
z
'x2
'y2
'z2
(1.37)
where
kx, ky, kz 5 hydraulic conductivity of the soil in the x, y, and z directions, respectively
h 5 hydraulic head at point A (i.e., the head of water that a piezometer placed at
A would show with the downstream water level as datum, as shown in
Figure 1.11)
For a two-dimensional flow condition, as shown in Figure 1.11,
'2h
50
'2y
1.11 Steady-State Seepage
Water level
29
Piezometers
hmax
h
O
O
Flow line
B
D
C
A
Water level
D
x
Permeable
soil layer
y
Equipotential line
F
E
Rock
z
Figure 1.11 Steady-state seepage
so Eq. (1.37) takes the form
kx
'2h
'2h
1
k
50
z
'x2
'z2
(1.38)
If the soil is isotropic with respect to hydraulic conductivity, kx 5 kz 5 k, and
'2h
'2h
1
50
'x2
'z2
(1.39)
Equation (1.39), which is referred to as Laplace’s equation and is valid for confined flow,
represents two orthogonal sets of curves known as flow lines and equipotential lines. A
flow net is a combination of numerous equipotential lines and flow lines. A flow line is a
path that a water particle would follow in traveling from the upstream side to the downstream side. An equipotential line is a line along which water, in piezometers, would rise to
the same elevation. (See Figure 1.11.)
In drawing a flow net, you need to establish the boundary conditions. For example, in
Figure 1.11, the ground surfaces on the upstream (OOr) and downstream (DDr) sides are
equipotential lines. The base of the dam below the ground surface, OrBCD, is a flow line.
The top of the rock surface, EF, is also a flow line. Once the boundary conditions are established, a number of flow lines and equipotential lines are drawn by trial and error so that all
the flow elements in the net have the same length-to-width ratio (L>B). In most cases, L>B
is held to unity, that is, the flow elements are drawn as curvilinear “squares.” This method is
illustrated by the flow net shown in Figure 1.12. Note that all flow lines must intersect all
equipotential lines at right angles.
Once the flow net is drawn, the seepage (in unit time per unit length of the structure)
can be calculated as
q 5 khmax
Nf
Nd
n
(1.40)
30 Chapter 1: Geotechnical Properties of Soil
Water level
hmax
Water level
B
Permeable soil layer
kx kz
L
Rock
Figure 1.12 Flow net
where
Nf 5 number of flow channels
Nd 5 number of drops
n 5 width-to-length ratio of the flow elements in the flow net (B>L)
hmax 5 difference in water level between the upstream and downstream sides
The space between two consecutive flow lines is defined as a flow channel, and the space
between two consecutive equipotential lines is called a drop. In Figure 1.12,
Nf 5 2, Nd 5 7, and n 5 1. When square elements are drawn in a flow net,
q 5 khmax
1.12
Nf
Nd
(1.41)
Effective Stress
The total stress at a given point in a soil mass can be expressed as
s 5 sr 1 u
(1.42)
where
s 5 total stress
sr 5 effective stress
u 5 pore water pressure
The effective stress, sr, is the vertical component of forces at solid-to-solid contact points
over a unit cross-sectional area. Referring to Figure 1.13a, at point A
s 5 gh1 1 gsath2
u 5 h2gw
where
gw 5 unit weight of water
gsat 5 saturated unit weight of soil
1.12 Effective Stress
31
h
Water level
Unit weight = h1
h2
B
Groundwater level
h1
Water
Saturated
unit weight = sat
A
h2
Saturated unit
weight = sat
A
F2
F1
X
Flow of water
(a)
(b)
Figure 1.13 Calculation of effective stress
So
sr 5 (gh1 1 gsath2 ) 2 (h2gw )
5 gh1 1 h2 (gsat 2 gw )
5 gh1 1 grh2
(1.43)
where gr 5 effective or submerged unit weight of soil.
For the problem in Figure 1.13a, there was no seepage of water in the soil. Figure
1.13b shows a simple condition in a soil profile in which there is upward seepage. For this
case, at point A,
s 5 h1gw 1 h2gsat
and
u 5 (h1 1 h2 1 h)gw
Thus, from Eq. (1.42),
sr 5 s 2 u 5 (h1gw 1 h2gsat ) 2 (h1 1 h2 1 h)gw
5 h2 (gsat 2 gw ) 2 hgw 5 h2gr 2 hgw
or
sr 5 h2 ¢gr 2
h
g ≤ 5 h2 (gr 2 igw )
h2 w
(1.44)
Note in Eq. (1.44) that h>h2 is the hydraulic gradient i. If the hydraulic gradient is very
high, so that gr 2 igw becomes zero, the effective stress will become zero. In other words,
there is no contact stress between the soil particles, and the soil will break up. This situation is referred to as the quick condition, or failure by heave. So, for heave,
i 5 icr 5
gr
Gs 2 1
5
gw
11e
(1.45)
32
Chapter 1: Geotechnical Properties of Soil
where icr 5 critical hydraulic gradient.
For most sandy soils, icr ranges from 0.9 to 1.1, with an average of about unity.
1.13
Consolidation
In the field, when the stress on a saturated clay layer is increased—for example, by the
construction of a foundation—the pore water pressure in the clay will increase. Because
the hydraulic conductivity of clays is very small, some time will be required for the excess
pore water pressure to dissipate and the increase in stress to be transferred to the soil skeleton. According to Figure 1.14, if Ds is a surcharge at the ground surface over a very large
area, the increase in total stress at any depth of the clay layer will be equal to Ds.
However, at time t 5 0 (i.e., immediately after the stress is applied), the excess pore
water pressure at any depth Du will equal Ds, or
Du 5 Dhigw 5 Ds (at time t 5 0)
Hence, the increase in effective stress at time t 5 0 will be
Dsr 5 Ds 2 Du 5 0
Theoretically, at time t 5 `, when all the excess pore water pressure in the clay layer has
dissipated as a result of drainage into the sand layers,
Du 5 0
(at time t 5 ` )
Then the increase in effective stress in the clay layer is
Dsr 5 Ds 2 Du 5 Ds 2 0 5 Ds
This gradual increase in the effective stress in the clay layer will cause settlement over a
period of time and is referred to as consolidation.
Laboratory tests on undisturbed saturated clay specimens can be conducted (ASTM
Test Designation D-2435) to determine the consolidation settlement caused by various
incremental loadings. The test specimens are usually 63.5 mm in diameter and 25.4 mm in
Immediately after
loading:
time t = 0
hi
Groundwater table
Sand
Clay
Sand
Figure 1.14 Principles of consolidation
1.13 Consolidation
33
height. Specimens are placed inside a ring, with one porous stone at the top and one at the
bottom of the specimen (Figure 1.15a). A load on the specimen is then applied so that the
total vertical stress is equal to s. Settlement readings for the specimen are taken periodically for 24 hours. After that, the load on the specimen is doubled and more settlement readings are taken. At all times during the test, the specimen is kept under water. The procedure
is continued until the desired limit of stress on the clay specimen is reached.
Based on the laboratory tests, a graph can be plotted showing the variation of the void
ratio e at the end of consolidation against the corresponding vertical effective stress sr. (On
a semilogarithmic graph, e is plotted on the arithmetic scale and sr on the log scale.) The
nature of the variation of e against log sr for a clay specimen is shown in Figure 1.15b. After
the desired consolidation pressure has been reached, the specimen gradually can be
unloaded, which will result in the swelling of the specimen. The figure also shows the variation of the void ratio during the unloading period.
From the e–log sr curve shown in Figure 1.15b, three parameters necessary for calculating settlement in the field can be determined. They are preconsolidation pressure
(scr ), compression index (Cc ), and the swelling index (Cs ). The following are more
detailed descriptions for each of the parameters.
Dial gauge
Load
Water Level
Porous stone
Ring
Soil Specimen
Porous stone
(a)
2.3
c
2.2
O
2.1
D
A
Void ratio, e
2.0
C
1.9
(e1, 1 )
1.8
B
Slope Cc
1.7
(e3, 3 )
1.6
Slope Cs
1.5
(e4, 4 )
(e2, 2 )
1.4
10
100
Effective pressure, (kN/m2)
(b)
400
Figure 1.15 (a) Schematic
diagram of consolidation test
arrangement; (b) e–log sr
curve for a soft clay from East
St. Louis, Illinois (Note: At
the end of consolidation,
s 5 sr)
34
Chapter 1: Geotechnical Properties of Soil
Preconsolidation Pressure
The preconsolidation pressure, scr , is the maximum past effective overburden pressure to
which the soil specimen has been subjected. It can be determined by using a simple graphical procedure proposed by Casagrande (1936). The procedure involves five steps (see
Figure 1.15b):
a. Determine the point O on the e–log sr curve that has the sharpest curvature (i.e., the
smallest radius of curvature).
b. Draw a horizontal line OA.
c. Draw a line OB that is tangent to the e–log sr curve at O.
d. Draw a line OC that bisects the angle AOB.
e. Produce the straight-line portion of the e–log sr curve backwards to intersect OC. This
is point D. The pressure that corresponds to point D is the preconsolidation
pressure scr .
Natural soil deposits can be normally consolidated or overconsolidated (or preconsolidated). If the present effective overburden pressure sr 5 sor is equal to the preconsolidated
pressure scr the soil is normally consolidated. However, if sor , scr , the soil is overconsolidated.
Stas and Kulhawy (1984) correlated the preconsolidation pressure with liquidity
index in the following form:
scr
5 10(1.1121.62 LI)
pa
(1.46)
where
pa ⫽ atmospheric pressure (⬇100 kN/m2)
LI ⫽ liquidity index
A similar correlation has also been provided by Kulhawy and Mayne (1990), which
is based on the work of Wood (1983) as
scr 5 sor e 10c122.5LI21.25loga pa bd f
sor
(1.47)
where ␴o⬘ ⫽ in situ effective overburden pressure.
Nagaraj and Murthy (1985) gave a correlation between ␴c⬘ and the in situ effective
overburden pressure which can be expressed as
eo
1.122 2 a b 2 0.0463log sor (kN>m2 )
e
L
log sor (kN>m2 ) 5
0.188
where
␴o⬘ ⫽ in situ effective overburden pressure
eo ⫽ in situ void ratio
LL(%)
d Gs
eL ⫽ void ratio at liquid limit 5 c
100
Gs ⫽ specific gravity of soil solids
(1.48)
1.13 Consolidation
35
Compression Index
The compression index, Cc, is the slope of the straight-line portion (the latter part) of the
loading curve, or
Cc 5
e1 2 e2
e1 2 e2
5
log s2r 2 log s1r
s2r
log ¢ ≤
s1r
(1.49)
where e1 and e2 are the void ratios at the end of consolidation under effective stresses
s1r and s2r , respectively.
The compression index, as determined from the laboratory e–log sr curve, will be somewhat different from that encountered in the field. The primary reason is that the soil remolds
itself to some degree during the field exploration. The nature of variation of the e–log sr curve
in the field for a normally consolidated clay is shown in Figure 1.16. The curve, generally
referred to as the virgin compression curve, approximately intersects the laboratory curve at a
void ratio of 0.42eo (Terzaghi and Peck, 1967). Note that eo is the void ratio of the clay in the
field. Knowing the values of eo and scr , you can easily construct the virgin curve and calculate
its compression index by using Eq. (1.49).
The value of Cc can vary widely, depending on the soil. Skempton (1944) gave an
empirical correlation for the compression index in which
Cc 5 0.009(LL 2 10)
(1.50)
where LL 5 liquid limit .
Besides Skempton, several other investigators also have proposed correlations for
the compression index. Some of those are given here:
Rendon-Herrero (1983):
Cc 5 0.141G 1.2
s a
Void ratio, e
e0
0 c
Virgin
compression
curve,
Slope Cc
e1
e2
1 1 eo 2.38
b
Gs
Laboratory
consolidation
curve
0.42 e0
1
2
Pressure, (log scale)
Figure 1.16 Construction of virgin
compression curve for normally
consolidated clay
(1.51)
36 Chapter 1: Geotechnical Properties of Soil
Nagaraj and Murty (1985):
Cc 5 0.2343 c
LL(%)
d Gs
100
(1.52)
no
371.747 2 4.275no
(1.53)
Park and Koumoto (2004):
Cc 5
where no 5 in situ porosity of soil.
Wroth and Wood (1978):
Cc 5 0.5Gs a
PI(%)
b
100
(1.54)
If a typical value of Gs ⫽ 2.7 is used in Eq. (1.54), we obtain (Kulhawy and Mayne, 1990)
Cc 5
PI(%)
74
(1.55)
Swelling Index
The swelling index, Cs, is the slope of the unloading portion of the e–log sr curve. In
Figure 1.15b, it is defined as
Cs 5
e3 2 e4
(1.56)
s4r
log ¢ ≤
s3r
In most cases, the value of the swelling index is 41 to 15 of the compression index.
Following are some representative values of Cs>Cc for natural soil deposits:
Description of soil
Cs , Cc
Boston Blue clay
Chicago clay
New Orleans clay
St. Lawrence clay
0.24–0.33
0.15– 0.3
0.15–0.28
0.05–0.1
The swelling index is also referred to as the recompression index.
The determination of the swelling index is important in the estimation of consolidation settlement of overconsolidated clays. In the field, depending on the pressure
increase, an overconsolidated clay will follow an e–log sr path abc, as shown in
Figure 1.17. Note that point a, with coordinates sor and eo, corresponds to the field conditions before any increase in pressure. Point b corresponds to the preconsolidation
pressure (scr ) of the clay. Line ab is approximately parallel to the laboratory unloading
curve cd (Schmertmann, 1953). Hence, if you know eo, sor , scr , Cc, and Cs, you can easily construct the field consolidation curve.
Using the modified Cam clay model and Eq. (1.54), Kulhawy and Mayne (1990)
have shown that
1.14 Calculation of Primary Consolidation Settlement
Void
ratio, e
e0
Slope
Cs c
a
b
Laboratory
consolidation
curve
d
Slope
Cs
0.42 e0
0
Virgin
compression
curve,
Slope Cc
c
Pressure, (log scale)
Figure 1.17 Construction of field
consolidation curve for overconsolidated clay
PI(%)
370
Comparing Eqs. (1.55) and (1.57), we obtain
Cs 5
1
Cs < Cc
5
1.14
37
(1.57)
(1.58)
Calculation of Primary Consolidation Settlement
The one-dimensional primary consolidation settlement (caused by an additional load) of a
clay layer (Figure 1.18) having a thickness Hc may be calculated as
Sc 5
De
H
1 1 eo c
(1.59)
Added pressure Groundwater table
Sand
Clay
He
Initial void
ratio eo
Sand
Average
effective
pressure
before load
application
o
Figure 1.18 One-dimensional settlement calculation
38
Chapter 1: Geotechnical Properties of Soil
where
Sc 5 primary consolidation settlement
De 5 total change of void ratio caused by the additional load application
eo 5 void ratio of the clay before the application of load
For normally consolidated clay (that is, sor 5 scr )
De 5 Cc log
sor 1 Dsr
sor
(1.60)
where
sor 5 average effective vertical stress on the clay layer
Dsr 5 Ds (that is, added pressure)
Now, combining Eqs. (1.59) and (1.60) yields
Sc 5
CcHc
sor 1 Dsr
log
1 1 eo
sor
For overconsolidated clay with sor 1 Dsr < scr ,
sor 1 Dsr
De 5 Cs log
sor
(1.61)
(1.62)
Combining Eqs. (1.59) and (1.62) gives
Sc 5
Cs Hc
sor 1 Dsr
log
1 1 eo
sor
(1.63)
For overconsolidated clay, if sor , scr , sor 1 Dsr, then
De 5 De1 1 De2 5 Cs log
scr
sor 1 Dsr
1 Cc log
sor
scr
(1.64)
Now, combining Eqs. (1.59) and (1.64) yields
Sc 5
1.15
CsHc
scr
CcHc
sor 1 Dsr
log
1
log
1 1 eo
sor
1 1 eo
scr
(1.65)
Time Rate of Consolidation
In Section 1.13 (see Figure 1.14), we showed that consolidation is the result of the gradual dissipation of the excess pore water pressure from a clay layer. The dissipation of pore
water pressure, in turn, increases the effective stress, which induces settlement. Hence, to
estimate the degree of consolidation of a clay layer at some time t after the load is applied,
you need to know the rate of dissipation of the excess pore water pressure.
1.15 Time Rate of Consolidation
z
Ground
water table
h
39
u
z 2H
Sand
Clay
Hc 2H
zH
t t2
T T(2)
t t1
T T (1)
A
0
Sand
(a)
z0
(b)
Figure 1.19 (a) Derivation
of Eq. (1.68); (b) nature of
variation of Du with time
Figure 1.19 shows a clay layer of thickness Hc that has highly permeable sand layers at its top and bottom. Here, the excess pore water pressure at any point A at any time t
after the load is applied is Du 5 (Dh)gw. For a vertical drainage condition (that is, in the
direction of z only) from the clay layer, Terzaghi derived the differential equation
'(Du)
'2 (Du)
5 Cv
't
'z2
(1.66)
where Cv 5 coefficient of consolidation, defined by
Cv 5
k
5
mvgw
k
De
g
Dsr(1 1 eav ) w
(1.67)
in which
k 5 hydraulic conductivity of the clay
De 5 total change of void ratio caused by an effective stress increase of Dsr
eav 5 average void ratio during consolidation
mv 5 volume coefficient of compressibility 5 De>3Dsr(1 1 eav )4
Equation (1.66) can be solved to obtain Du as a function of time t with the following
boundary conditions:
1. Because highly permeable sand layers are located at z 5 0 and z 5 Hc, the excess
pore water pressure developed in the clay at those points will be immediately dissipated. Hence,
Du 5 0
at z 5 0
and
Du 5 0 at z 5 Hc 5 2H
where H 5 length of maximum drainage path (due to two-way drainage
condition—that is, at the top and bottom of the clay).
40 Chapter 1: Geotechnical Properties of Soil
2. At time t 5 0, Du 5 Du0 5 initial excess pore water pressure after the load is
applied. With the preceding boundary conditions, Eq. (1.66) yields
m5 `
2(Du0 )
Mz
2
sin ¢
≤ Re 2M Tv
M
H
Du 5 a B
m50
(1.68)
where
M 5 3(2m 1 1)p4>2
m 5 an integer 5 1, 2, c
Tv 5 nondimensional time factor 5 (Cv t)>H 2
(1.69)
The value of Du for various depths (i.e., z 5 0 to z 5 2H) at any given time t (and thus Tv)
can be calculated from Eq. (1.68). The nature of this variation of Du is shown in Figures 1.20a and b. Figure 1.20c shows the variation of Du>Du0 with Tv and H>Hc using
Eqs.(1.68) and (1.69).
The average degree of consolidation of the clay layer can be defined as
U5
Sc(t)
(1.70)
Sc(max)
where
Sc(t) 5 settlement of a clay layer at time t after the load is applied
Sc(max) 5 maximum consolidation settlement that the clay will undergo under a given
loading
If the initial pore water pressure (Du0 ) distribution is constant with depth, as shown
in Figure 1.20a, the average degree of consolidation also can be expressed as
2H
U5
Sc(t)
Sc(max)
2H
3 (Du0 )dz 2 3 ( Du)dz
5
0
0
(1.71)
2H
3 (Du0 )dz
0
or
2H
(Du0 )2H 2 3
2H
0
U5
3
(Du)dz
(Du0 )2H
512
(Du)dz
0
2H(Du0 )
(1.72)
Now, combining Eqs. (1.68) and (1.72), we obtain
U5
Sc(t)
Sc(max)
m5 `
2
2
5 1 2 a ¢ 2 ≤ e 2M Tv
M
m50
(1.73)
The variation of U with Tv can be calculated from Eq. (1.73) and is plotted in Figure
1.21. Note that Eq. (1.73) and thus Figure 1.21 are also valid when an impermeable layer
is located at the bottom of the clay layer (Figure 1.20). In that case, the dissipation of
1.15 Time Rate of Consolidation
Highly permeable
layer (sand)
Highly permeable
layer (sand)
u at
t0
u at
t0
u0 Hc 2H
41
u0 Hc H
constant
with
depth
Highly permeable
layer (sand)
(a)
constant
with
depth
Impermeable
layer
(b)
2.0
T 0
1.5
H
Hc
T 1
0.9
1.0
0.6 0.5
0.7
0.4
0.3
0.2
T 0.1
0.8
0.5
0
0
0.1
0.2
0.3 0.4 0.5 0.6 0.7 0.8
Excess pore water pressure, u
Initial excess pore water pressure, u0
0.9
1.0
(c)
Figure 1.20 Drainage condition for consolidation: (a) two-way drainage; (b) one-way drainage;
(c) plot of Du>Du0 with Tv and H>Hc
excess pore water pressure can take place in one direction only. The length of the maximum
drainage path is then equal to H 5 Hc.
The variation of Tv with U shown in Figure 1.21 can also be approximated by
Tv 5
p U% 2
¢
≤
4 100
(for U 5 0 to 60%)
(1.74)
and
Tv 5 1.781 2 0.933 log (100 2 U%)
(for U . 60%)
(1.75)
42 Chapter 1: Geotechnical Properties of Soil
Eq (1.74)
Eq (1.75)
1.0
Sand
Sand
Time factor, T
0.8
2H = Hc
H = Hc
Clay
Clay
0.6
Sand u0 = constant
0.4
Rock u0 = constant
0.2
0
0
10
20
30
40
50
60
70
Average degree of consolidation, U (%)
80
90
Figure 1.21 Plot of time factor against average degree of consolidation (Du0 5 constant)
Table 1.11 gives the variation of Tv with U on the basis of Eqs. (1.74) and (1.75)
Sivaram and Swamee (1977) gave the following equation for U varying from 0 to
100%:
(4Tv>p) 0.5
U%
5
100
31 1 (4Tv>p) 2.84 0.179
or
Tv 5
(1.76)
(p>4) (U%>100) 2
31 2 (U%>100) 5.64 0.357
(1.77)
Equations (1.76) and (1.77) give an error in Tv of less than 1% for 0% ⬍ U ⬍ 90% and
less than 3% for 90% ⬍ U ⬍ 100%.
Table 1.11 Variation of Tv with U
U (%)
0
1
2
3
4
5
6
7
8
9
10
Tv
0
0.00008
0.0003
0.00071
0.00126
0.00196
0.00283
0.00385
0.00502
0.00636
0.00785
U (%)
Tv
U (%)
Tv
U (%)
Tv
26
27
28
29
30
31
32
33
34
35
36
0.0531
0.0572
0.0615
0.0660
0.0707
0.0754
0.0803
0.0855
0.0907
0.0962
0.102
52
53
54
55
56
57
58
59
60
61
62
0.212
0.221
0.230
0.239
0.248
0.257
0.267
0.276
0.286
0.297
0.307
78
79
80
81
82
83
84
85
86
87
88
0.529
0.547
0.567
0.588
0.610
0.633
0.658
0.684
0.712
0.742
0.774
1.15 Time Rate of Consolidation
43
Table 1.11 (Continued)
U (%)
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Tv
U (%)
Tv
U (%)
Tv
U (%)
Tv
0.0095
0.0113
0.0133
0.0154
0.0177
0.0201
0.0227
0.0254
0.0283
0.0314
0.0346
0.0380
0.0415
0.0452
0.0491
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
0.107
0.113
0.119
0.126
0.132
0.138
0.145
0.152
0.159
0.166
0.173
0.181
0.188
0.197
0.204
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
0.318
0.329
0.304
0.352
0.364
0.377
0.390
0.403
0.417
0.431
0.446
0.461
0.477
0.493
0.511
89
90
91
92
93
94
95
96
97
98
99
100
0.809
0.848
0.891
0.938
0.993
1.055
1.129
1.219
1.336
1.500
1.781
⬁
Example 1.9
A laboratory consolidation test on a normally consolidated clay showed the following
results:
Load, Ds9 (kN , m2)
Void ratio at the
end of consolidation, e
140
212
0.92
0.86
The specimen tested was 25.4 mm in thickness and drained on both sides. The time required for the specimen to reach 50% consolidation was 4.5 min.
A similar clay layer in the field 2.8 m thick and drained on both sides, is subjected
to a similar increase in average effective pressure (i.e., s0r 5 140 kN>m2 and
s0r 1 Dsr 5 212 kN>m2 ). Determine
a. the expected maximum primary consolidation settlement in the field.
b. the length of time required for the total settlement in the field to reach 40 mm.
(Assume a uniform initial increase in excess pore water pressure with depth.)
Solution
Part a
For normally consolidated clay [Eq. (1.49)],
Cc 5
e1 2 e2
s2r
log ¢ ≤
s1r
5
0.92 2 0.86
212
log ¢
≤
140
5 0.333
44 Chapter 1: Geotechnical Properties of Soil
From Eq. (1.61),
Sc 5
CcHc
s0r 1 Dsr
(0.333) (2.8)
212
log
5
log
5 0.0875 m 5 87.5 mm
1 1 e0
s0r
1 1 0.92
140
Part b
From Eq. (1.70), the average degree of consolidation is
U5
Sc(t)
Sc(max)
5
40
(100) 5 45.7%
87.5
The coefficient of consolidation, Cv, can be calculated from the laboratory test.
From Eq. (1.69),
Tv 5
Cv t
H2
For 50% consolidation (Figure 1.21), Tv 5 0.197, t 5 4.5 min, and H 5 Hc>2 5
12.7 mm, so
Cv 5 T50
(0.197) (12.7) 2
H2
5
5 7.061 mm2>min
t
4.5
Again, for field consolidation, U 5 45.7%. From Eq. (1.74)
Tv 5
p U% 2 p 45.7 2
¢
≤ 5 ¢
≤ 5 0.164
4 100
4 100
But
Tv 5
Cv t
H2
or
2
t5
1.16
TvH 2
5
Cv
2.8 3 1000
0.164¢
≤
2
7.061
5 45,523 min 5 31.6 days
■
Degree of Consolidation Under Ramp Loading
The relationships derived for the average degree of consolidation in Section 1.15 assume
that the surcharge load per unit area (Ds) is applied instantly at time t 5 0. However, in
most practical situations, Ds increases gradually with time to a maximum value and
remains constant thereafter. Figure 1.22 shows Ds increasing linearly with time (t) up to
a maximum at time tc (a condition called ramp loading). For t $ tc, the magnitude of Ds
1.16 Degree of Consolidation Under Ramp Loading
z
45
Sand
Clay
2H = Hc
Sand
(a)
Load per unit
area, tc
(b)
Time, t
Figure 1.22 One-dimensional consolidation due to single
ramp loading
remains constant. Olson (1977) considered this phenomenon and presented the average
degree of consolidation, U, in the following form:
For Tv # Tc,
U5
Tv
2 m5 ` 1
2
b1 2
a M 4 31 2 exp(2M Tv )4 r
Tc
Tv m50
(1.78)
and for Tv $ Tc,
U512
2 m5 ` 1
2
2
a M 4 3exp(M Tc ) 2 14exp(2M Tc )
Tc m50
(1.79)
where m, M, and Tv have the same definition as in Eq. (1.68) and where
Tc 5
Cv tc
H2
(1.80)
Figure 1.23 shows the variation of U with Tv for various values of Tc, based on the
solution given by Eqs. (1.78) and (1.79).
Chapter 1: Geotechnical Properties of Soil
0
20
Tc 10
0.01
0.04
1 2
0.2
0.1
5
0.5
40
U (%)
46
60
80
100
0.01
0.1
1.0
10
Time factor, T
Figure 1.23 Olson’s ramp-loading solution: plot of U versus Tv (Eqs. 1.78 and 1.79)
Example 1.10
In Example 1.9, Part (b), if the increase in Ds would have been in the manner shown in
Figure 1.24, calculate the settlement of the clay layer at time t 5 31.6 days after the beginning of the surcharge.
Solution
From Part (b) of Example 1.9, Cn 5 7.061 mm2>min. From Eq. (1.80),
Tc 5
Cv tc
H2
5
(7.061 mm2>min) (15 3 24 3 60 min)
2
2.8
3 1000 mm≤
¢
2
5 0.0778
72 kN/m2
tc 15 days
Time, t
Figure 1.24 Ramp loading
1.17 Shear Strength
47
Also,
Tv 5
Cv t
H2
5
(7.061 mm2>min) (31.6 3 24 3 60 min)
2
2.8
¢
3 1000 mm ≤
2
5 0.164
From Figure 1.23, for Tv 5 0.164 and Tc 5 0.0778, the value of U is about 36%. Thus,
Sc(t531.6 days) 5 Sc(max) (0.36) 5 (87.5) (0.36) 5 31.5 mm
1.17
■
Shear Strength
The shear strength of a soil, defined in terms of effective stress, is
s 5 cr 1 sr tan fr
(1.81)
where
sr 5 effective normal stress on plane of shearing
cr 5 cohesion, or apparent cohesion
fr 5 effective stress angle of friction
Equation (1.81) is referred to as the Mohr–Coulomb failure criterion. The value of
cr for sands and normally consolidated clays is equal to zero. For overconsolidated clays,
cr . 0.
For most day-to-day work, the shear strength parameters of a soil (i.e., cr and fr) are
determined by two standard laboratory tests: the direct shear test and the triaxial test.
Direct Shear Test
Dry sand can be conveniently tested by direct shear tests. The sand is placed in a shear box
that is split into two halves (Figure 1.25a). First a normal load is applied to the specimen.
Then a shear force is applied to the top half of the shear box to cause failure in the sand.
The normal and shear stresses at failure are
sr 5
N
A
s5
R
A
and
where A 5 area of the failure plane in soil—that is, the cross-sectional area of the
shear box.
48 Chapter 1: Geotechnical Properties of Soil
Shear
stress
N
sc
s4
tan s3
R
s2
s1
1
(a)
2
3
4
(b)
Effective
normal
stress,
Figure 1.25 Direct shear test in sand: (a) schematic diagram of test equipment;
(b) plot of test results to obtain the friction angle fr
Several tests of this type can be conducted by varying the normal load. The angle of
friction of the sand can be determined by plotting a graph of s against sr (5 s for dry
sand), as shown in Figure 1.25b, or
fr 5 tan 21 ¢
s
≤
sr
(1.82)
For sands, the angle of friction usually ranges from 26° to 45°, increasing with the
relative density of compaction. A general range of the friction angle, fr, for sands is given
in Table 1.12.
In 1970, Brinch Hansen (see Hansbo, 1975, and Thinh, 2001) gave the following
correlation for ␾⬘ of granular soils.
␾⬘ (deg) ⫽ 26° + 10Dr + 0.4Cu + 1.6 log (D50)
(1.83)
where
Dr ⫽ relative density (fraction)
Cu ⫽ uniformity coefficient
D50 ⫽ mean grain size, in mm (i.e., the diameter through which 50% of the soil passes)
Table 1.12 Relationship between Relative Density and Angle of Friction of
Cohesionless Soils
State of packing
Very loose
Loose
Compact
Dense
Very dense
Relative
density
(%)
Angle of
friction, f9 (deg.)
,20
20–40
40–60
60–80
.80
,30
30–35
35–40
40–45
.45
1.17 Shear Strength
49
Teferra (1975) suggested the following empirical correlation based on a large data
base.
f9 (deg) 5 tan21 a
1
b
ae1b
(1.84)
where
e ⫽ void ratio
a 5 2.101 1 0.097a
D85
b
D15
(1.85)
b ⫽ 0.845 ⫺ 0.398a
(1.86)
D85 and D15 ⫽ diameters through which, respectively, 85% and 15% of soil passes
Thinh (2001) suggested that Eq. (1.84) provides as better correlation for ␾⬘ compared to
Eq. (1.83).
Triaxial Tests
Triaxial compression tests can be conducted on sands and clays Figure 1.26a shows a
schematic diagram of the triaxial test arrangement. Essentially, the test consists of placing
a soil specimen confined by a rubber membrane into a lucite chamber and then applying
an all-around confining pressure (s3 ) to the specimen by means of the chamber fluid
(generally, water or glycerin). An added stress (Ds) can also be applied to the specimen
in the axial direction to cause failure ( Ds 5 Dsf at failure). Drainage from the specimen
can be allowed or stopped, depending on the condition being tested. For clays, three main
types of tests can be conducted with triaxial equipment (see Figure 1.27):
1. Consolidated-drained test (CD test)
2. Consolidated-undrained test (CU test)
3. Unconsolidated-undrained test (UU test)
Consolidated-Drained Tests:
Step 1. Apply chamber pressure s3. Allow complete drainage, so that the pore water pressure (u 5 u0 ) developed is zero.
Step 2. Apply a deviator stress Ds slowly. Allow drainage, so that the pore water
pressure (u 5 ud ) developed through the application of Ds is zero. At failure, Ds 5 Dsf; the total pore water pressure uf 5 u0 1 ud 5 0.
So for consolidated-drained tests, at failure,
Major principal effective stress 5 s3 1 Dsf 5 s1 5 s1r
Minor principal effective stress 5 s3 5 s3r
Changing s3 allows several tests of this type to be conducted on various clay specimens. The
shear strength parameters (cr and fr) can now be determined by plotting Mohr’s circle at failure, as shown in Figure 1.26b, and drawing a common tangent to the Mohr’s circles. This is the
Mohr–Coulomb failure envelope. (Note: For normally consolidated clay, cr < 0.) At failure,
s1r 5 s3r tan2 ¢ 45 1
fr
fr
≤ 1 2cr tan ¢45 1 ≤
2
2
(1.87)
Piston
Porous
stone
Lucite
chamber
Rubber
membrane
Chamber
fluid
Soil
specimen
Shear
stress
Porous
stone
Base
plate
Valve
To drainage and/or
pore water pressure
device
Chamber
fluid
1
c
3
3
1
Consolidated-drained test
Schematic diagram of triaxial
test equipment
Effective
normal
stress
(b)
(a)
Shear
stress
Shear
stress
Total stress
failure
envelope
Effective
stress
failure
envelope
c
3
3
1
1
c
Total
normal
stress,
3
3
1
Consolidated-undrained test
(c)
Shear
stress
Total stress
failure envelope
( 0)
s cu
1
3
3
1
Unconsolidated-undrained test
(d)
Figure 1.26 Triaxial test
Normal stress
(total),
1
Effective
normal
stress,
1.17 Shear Strength
51
3
3
3
3
3
3
3
3
Figure 1.27 Sequence of stress
application in triaxial test
Consolidated-Undrained Tests:
Step 1. Apply chamber pressure s3. Allow complete drainage, so that the pore water pressure (u 5 u0 ) developed is zero.
Step 2. Apply a deviator stress Ds. Do not allow drainage, so that the pore water
pressure u 5 ud 2 0. At failure, Ds 5 Dsf; the pore water pressure
uf 5 u0 1 ud 5 0 1 ud(f).
Hence, at failure,
Major principal total stress 5 s3 1 Dsf 5 s1
Minor principal total stress 5 s3
Major principal effective stress 5 (s3 1 Dsf ) 2 uf 5 s1r
Minor principal effective stress 5 s3 2 uf 5 s3r
Changing s3 permits multiple tests of this type to be conducted on several soil specimens. The total stress Mohr’s circles at failure can now be plotted, as shown in
Figure 1.26c, and then a common tangent can be drawn to define the failure envelope. This
total stress failure envelope is defined by the equation
s 5 c 1 s tan f
(1.88)
where c and f are the consolidated-undrained cohesion and angle of friction, respectively.
(Note: c < 0 for normally consolidated clays.)
Similarly, effective stress Mohr’s circles at failure can be drawn to determine the effective stress failure envelope (Figure 1.26c), which satisfy the relation expressed in Eq. (1.81).
Unconsolidated-Undrained Tests:
Step 1. Apply chamber pressure s3. Do not allow drainage, so that the pore water
pressure (u 5 u0 ) developed through the application of s3 is not zero.
Step 2. Apply a deviator stress Ds. Do not allow drainage (u 5 ud 2 0). At failure,
Ds 5 Dsf; the pore water pressure uf 5 u0 1 ud(f)
For unconsolidated-undrained triaxial tests,
Major principal total stress 5 s3 1 Dsf 5 s1
Minor principal total stress 5 s3
52 Chapter 1: Geotechnical Properties of Soil
The total stress Mohr’s circle at failure can now be drawn, as shown in Figure 1.26d.
For saturated clays, the value of s1 2 s3 5 Dsf is a constant, irrespective of the chamber
confining pressure s3 (also shown in Figure 1.26d). The tangent to these Mohr’s circles will
be a horizontal line, called the f 5 0 condition. The shear strength for this condition is
s 5 cu 5
Dsf
(1.89)
2
where cu 5 undrained cohesion (or undrained shear strength).
The pore pressure developed in the soil specimen during the unconsolidatedundrained triaxial test is
u 5 u0 1 ud
(1.90)
The pore pressure u0 is the contribution of the hydrostatic chamber pressure s3. Hence,
u0 5 Bs3
(1.91)
where B 5 Skempton’s pore pressure parameter.
Similarly, the pore parameter ud is the result of the added axial stress Ds, so
ud 5 ADs
(1.92)
where A 5 Skempton’s pore pressure parameter.
However,
Ds 5 s1 2 s3
(1.93)
Combining Eqs. (1.90), (1.91), (1.92), and (1.93) gives
u 5 u0 1 ud 5 Bs3 1 A(s1 2 s3 )
(1.94)
The pore water pressure parameter B in soft saturated soils is approximately 1, so
u 5 s3 1 A(s1 2 s3 )
(1.95)
The value of the pore water pressure parameter A at failure will vary with the type of soil.
Following is a general range of the values of A at failure for various types of clayey soil encountered in nature:
Type of soil
Sandy clays
Normally consolidated clays
Overconsolidated clays
1.18
A at failure
0.5–0.7
0.5–1
20.5– 0
Unconfined Compression Test
The unconfined compression test (Figure 1.28a) is a special type of unconsolidatedundrained triaxial test in which the confining pressure s3 5 0, as shown in
Figure 1.28b. In this test, an axial stress Ds is applied to the specimen to cause failure
1.18 Unconfined Compression Test
53
Specimen
(a)
Unconfined
compression
strength, qu
Shear
stress
cu
3 0
1 ƒ
qu
Total
normal
stress
Degree
of
saturation
(b)
(c)
Figure 1.28 Unconfined compression test: (a) soil specimen; (b) Mohr’s circle for the
test; (c) variation of qu with the degree of saturation
(i.e., Ds 5 Dsf ). The corresponding Mohr’s circle is shown in Figure 1.28b. Note that,
for this case,
Major principal total stress 5 Dsf 5 qu
Minor principal total stress 5 0
The axial stress at failure, Dsf 5 qu, is generally referred to as the unconfined compression strength. The shear strength of saturated clays under this condition (f 5 0),
from Eq. (1.81), is
s 5 cu 5
qu
2
(1.96)
The unconfined compression strength can be used as an indicator of the consistency of
clays.
Unconfined compression tests are sometimes conducted on unsaturated soils. With the
void ratio of a soil specimen remaining constant, the unconfined compression strength rapidly
decreases with the degree of saturation (Figure 1.28c).
54 Chapter 1: Geotechnical Properties of Soil
Comments on Friction Angle, f9
Effective Stress Friction Angle of Granular Soils
In general, the direct shear test yields a higher angle of friction compared with that obtained
by the triaxial test. Also, note that the failure envelope for a given soil is actually curved. The
Mohr–Coulomb failure criterion defined by Eq. (1.81) is only an approximation. Because of
the curved nature of the failure envelope, a soil tested at higher normal stress will yield a lower
value of fr. An example of this relationship is shown in Figure 1.29, which is a plot of fr
versus the void ratio e for Chattachoochee River sand near Atlanta, Georgia (Vesic, 1963). The
friction angles shown were obtained from triaxial tests. Note that, for a given value of e, the
magnitude of fr is about 4° to 5° smaller when the confining pressure s3r is greater than about
70 kN>m2 , compared with that when s3r , 70 kN>m2.
Effective Stress Friction Angle of Cohesive Soils
Figure 1.30 shows the variation of effective stress friction angle, f9, for several normally consolidated clays (Bjerrum and Simons, 1960; Kenney, 1959). It can be seen from the figure that,
in general, the friction angle fr decreases with the increase in plasticity index. The value of
fr generally decreases from about 37 to 38° with a plasticity index of about 10 to about 25°
or less with a plasticity index of about 100. The consolidated undrained friction angle (f) of
normally consolidated saturated clays generally ranges from 5 to 20°.
45
7 samples
6 samples
Effective stress friction angle, (deg)
1.19
e tan = 0.68 [ 3 < 70 kN/m2]
40
8
samples
6 samples
5 samples
10 samples
7 samples
35
7 samples
e tan = 0.59 [70 kN/m
< 3 < 550 kN/m2
2]
30
0.6
0.7
0.8
0.9
Void ratio, e
1.0
1.1
1.2
Figure 1.29 Variation of friction angle fr with void ratio for Chattachoochee River sand (After
Vesic, 1963) (From Vesic, A. B. Bearing Capacity of Deep Foundations in Sand. In Highway
Research Record 39, Highway Research Board, National Research Council, Washington, D.C.,
1963, Figure 11, p. 123. Reproduced with permission of the Transportation Research Board.)
1.19 Comments on Friction Angle, fr
55
1.0
Kenney (1959)
Sin 0.8
Bjerrum and
Simons (1960)
0.6
0.4
0.2
0
5
10
20
30
50
Plasticity index (%)
80 100
150
Figure 1.30 Variation of sin fr with plasticity index (PI) for several normally
consolidated clays
The consolidated drained triaxial test was described in Section 1.17. Figure 1.31
shows a schematic diagram of a plot of Ds versus axial strain in a drained triaxial test for
a clay. At failure, for this test, Ds 5 Dsf. However, at large axial strain (i.e., the ultimate
strength condition), we have the following relationships:
Major principal stress: s1(ult)
r
5 s3 1 Dsult
Minor principal stress: s3(ult)
r
5 s3
At failure (i.e., peak strength), the relationship between s1r and s3r is given by Eq.
(1.87). However, for ultimate strength, it can be shown that
s1(ult)
r
5 s3r tan2 ¢ 45 1
frr
≤
2
(1.97)
where frr 5 residual effective stress friction angle.
Figure 1.32 shows the general nature of the failure envelopes at peak strength and
ultimate strength (or residual strength). The residual shear strength of clays is important
in the evaluation of the long-term stability of new and existing slopes and the design of
remedial measures. The effective stress residual friction angles frr of clays may be substantially smaller than the effective stress peak friction angle fr. Past research has shown
that the clay fraction (i.e., the percent finer than 2 microns) present in a given soil, CF, and
Deviator
stress, f
ult
3 = 3 = constant
Axial strain,
Figure 1.31 Plot of deviator stress versus
axial strain–drained triaxial test
Chapter 1: Geotechnical Properties of Soil
Shear stress, c
ed
an dat
t onsoli
+
c
c
ver
s=
—o
h
ed
t
ng
dat
stre
n onsoli
a
t
k
a
c
Pe
ally
s=
orm
n
—
gth
tren
s
tan r
k
s=
Pea
gth
al stren
Residu
r
Effective normal stress, Figure 1.32 Peak- and residual-strength envelopes for clay
the clay mineralogy are the two primary factors that control frr. The following is a summary of the effects of CF on frr.
1. If CF is less than about 15%, then frr is greater than about 25°.
2. For CF . about 50%, frr is entirely governed by the sliding of clay minerals and
may be in the range of about 10 to 15°.
3. For kaolinite, illite, and montmorillonite, frr is about 15°, 10°, and 5°, respectively.
Illustrating these facts, Figure 1.33 shows the variation of frr with CF for several soils
(Skempton, 1985).
40
Skempton (1985)
pa ≈ 1
Sand
Residual friction angle, r (deg)
56
30
Plasticity index, PI
= 0.5 to 0.9
Clay fraction, CF
20
Kaolin
10
Bentonite
0
0
20
40
60
Clay fraction, CF (%)
80
Figure 1.33 Variation of frr with CF (Note: pa 5 atmospheric pressure)
100
1.21 Sensitivity
1.20
57
Correlations for Undrained Shear Strength, Cu
Several empirical relationships can be observed between cu and the effective overburden
pressure (␴0⬘) in the field. Some of these relationships are summarized in Table 1.13.
Table 1.13 Empirical Equations Related to cu and ␴0⬘
Reference
Relationship
Remarks
cu(VST)
5 0.11 1 0.00037 (PI)
s0r
PI ⫽ plasticity index (%)
cu(VST) ⫽ undrained shear
strength from vane shear test
cu(VST)
5 0.11 1 0.0037 (PI)
scr
Skempton (1957)
Chandler (1988)
Jamiolkowski, et al. (1985)
Mesri (1989)
Bjerrum and Simons (1960)
For normally consolidated clay
Can be used in overconsolidated
soil; accuracy ⫾25%; not valid
for sensitive and fissured clays
␴c⬘ ⫽ preconsolidation pressure
cu
5 0.23 6 0.04
scr
cu
5 0.22
s0r
cu
PI% 0.5
5 0.45a
b
s0r
100
For lightly overconsolidated clays
Normally consolidated clay
for PI ⬎ 50%
cu
5 0.118 (LI) 0.15
s0r
Normally consolidated clay
for LI ⫽ liquidity index ⬎ 0.5
Ladd, et al. (1977)
a
cu
b
s0r overconsolidated
cu
a b
sr0 normally consolidated
5 OCR0.8
OCR ⫽ overconsolidation ratio ⫽ ␴c⬘/␴0⬘
1.21
Sensitivity
For many naturally deposited clay soils, the unconfined compression strength is much less
when the soils are tested after remolding without any change in the moisture content. This
property of clay soil is called sensitivity. The degree of sensitivity is the ratio of the unconfined compression strength in an undisturbed state to that in a remolded state, or
St 5
qu(undisturbed)
qu(remolded)
(1.98)
58 Chapter 1: Geotechnical Properties of Soil
The sensitivity ratio of most clays ranges from about 1 to 8; however, highly flocculent marine clay deposits may have sensitivity ratios ranging from about 10 to 80. Some clays
turn to viscous liquids upon remolding, and these clays are referred to as “quick” clays. The
loss of strength of clay soils from remolding is caused primarily by the destruction of the clay
particle structure that was developed during the original process of sedimentation.
Problems
1.1
1.2
1.3
1.4
1.5
1.6
A moist soil has a void ratio of 0.65. The moisture content of the soil is 14% and
Gs ⫽ 2.7. Determine:
a. Porosity
b. Degree of saturation (%)
c. Dry unit weight (kN/m3)
For the soil described in Problem 1.1:
a. What would be the saturated unit weight in kN/m3?
b. How much water, in kN/m3, needs to be added to the soil for complete saturation?
c. What would be the moist unit weight in kN/m3 when the degree of saturation
is 70%?
The moist unit weight of a soil is 18.79 kN/m3. For a moisture content of 12% and
Gs ⫽ 2.65, calculate:
a. Void ratio
b. Porosity
c. Degree of saturation
d. Dry unit weight
A saturated soil specimen has w ⫽ 36% and ␥d ⫽ 13.43 kN/m3. Determine:
a. Void ratio
b. Porosity
c. Specific gravity of soil solids
d. Saturated unit weight (kN/m3)
The laboratory test results of a sand are emax ⫽ 0.91, emin ⫽ 0.48, and Gs ⫽ 2.67.
What would be the dry and moist unit weights of this sand when compacted at a
moisture content of 10% to a relative density of 65%?
The laboratory test results of six soils are given in the following table. Classify the
soils by the AASHTO soil classification system and give the group indices.
Sieve Analysis—Percent Passing
1.7
1.8
Sieve No.
Soil A
Soil B
Soil C
Soil D
Soil E
Soil F
4
10
40
200
Liquid limit
Plastic limit
92
48
28
13
31
26
100
60
41
33
38
25
100
98
82
72
56
31
95
90
79
64
35
26
100
91
80
30
43
29
100
82
74
55
35
21
Classify the soils in Problem 1.6 using the Unified Soil Classification System. Give
group symbols and group names.
The permeability of a sand was tested in the laboratory at a void ratio of 0.6 and was
determined to be 0.14 cm/sec. Use Eq. (1.32) to estimate the hydraulic conductivity
of this sand at a void ratio of 0.8.
Problems
59
A sand has the following: D10 ⫽ 0.08 mm; D60 ⫽ 0.37 mm; void ratio e ⫽ 0.6.
a. Determine the hydraulic conductivity using Eq. (1.33).
b. Determine the hydraulic conductivity using Eq. (1.35).
1.10 The in situ hydraulic conductivity of a normally consolidated clay is 5.4 ⫻ 10⫺6 cm/sec
at a void ratio of 0.92. What would be its hydraulic conductivity at a void ratio of 0.72?
Use Eq. (1.36) and n ⫽ 5.1.
1.11 Refer to the soil profile shown in Figure P1.11. Determine the total stress, pore water
pressure, and effective stress at A, B, C, and D.
1.12 For a normally consolidated clay layer, the following are given:
1.9
Thickness ⫽ 3 m
Void ratio ⫽ 0.75
Liquid limit ⫽ 42
Gs ⫽ 2.72
Average effective stress on the clay layer ⫽ 110 kN/m2
How much consolidation settlement would the clay undergo if the average effective
stress on the clay layer is increased to 155 kN/m2 as the result of the construction of
a foundation? Use Eq. (1.51) to estimate the compression index.
1.13 Refer to Problem 1.12. Assume that the clay layer is preconsolidated, ␴c⬘ ⫽ 130 kN/m2,
and CS 5 15Cc . Estimate the consolidation settlement.
1.14 A saturated clay deposit below the ground water table in the field has liquid limit ⫽
61%, plastic limit ⫽ 21%, and moisture content ⫽ 38%. Estimate the preconsolidation pressure, ␴c⬘ (kN/m2) using Eq. (1.46).
1.15 A normally consolidated clay layer in the field has a thickness of 3.2 m with an average effective stress of 70 kN/m2. A laboratory consolidation test on the clay gave the
following results.
Pressure (kN , m2)
100
200
Void ratio
0.905
0.815
a. Determine the compression index, Cc.
b. If the average effective stress on the clay layer (␴o⬘ + ⌬␴) is increased to
115 kN/m2, what would be the total consolidation settlement?
A
Dry sand; e = 0.55 Gs = 2.66
3m
B
1.5 m
Water table
Sand Gs = 2.66 e = 0.48
C
5m
Clay w = 34.78% Gs = 2.74
D
Rock
Figure P1.11
60 Chapter 1: Geotechnical Properties of Soil
1.16 A clay soil specimen, 25.4 mm thick (drained on top and bottom), was tested in the laboratory. For a given load increment, the time for 50% consolidation was 5 min 20 sec.
How long will it take for 50% consolidation of a similar clay layer in the field that is
2.5 m thick and drained on one side only?
1.17 A clay soil specimen 25.4 mm thick (drained on top only) was tested in the laboratory. For a given load increment, the time for 60% consolidation was 6 min 20 sec.
How long will it take for 50% consolidation for a similar clay layer in the field that is
3.05 m thick and drained on both sides?
1.18 Refer to Figure P1.18. A total of 60-mm consolidation settlement is expected in the
two clay layers due to a surcharge ⌬␴. Find the duration of surcharge application at
which 30 mm of total settlement would take place.
1.19 The coefficient of consolidation of a clay for a given pressure range was obtained as
8 ⫻ 10⫺3 mm2/sec on the basis of one-dimensional consolidation test results. In the
field, there is a 2-m thick layer of the same clay (Figure P1.19a). Based on the assumption that a uniform surcharge of 70 kN/m2 was to be applied instantaneously,
the total consolidation settlement was estimated to be 150 mm.
However, during construction, the loading was gradual; the resulting surcharge
can be approximated as shown in Figure P1.19b. Estimate the settlement at t ⫽ 30
days and t ⫽ 120 days after the beginning of construction.
1.20 A direct shear test was conducted on a 2 ⫻ 2 specimen of dry sand had the following
results.
Normal force (N)
146.8
245.4
294.3
Shear force at failure (N)
91.9
159.2
178.8
Draw a graph of shear stress at failure versus normal stress and determine the soil
friction angle.
1m
Groundwater table
1m
Sand
2m
Clay
C = 2 mm2/min
1m
Sand
1m
Clay
C = 2 mm2/min
Sand
Figure P1.18
Problems
61
(kN/m2)
Sand
1m
Clay
2m
70 kN/m2
Sand
60
(a)
Time, days
(b)
Figure P1.19
1.21 For a sand, given:
D85 ⫽ 0.21 mm
D50 ⫽ 0.13 mm
D15 ⫽ 0.09 mm
Uniformity coefficient, Cu ⫽ 2.1
Void ratio, e ⫽ 0.68
Relative density ⫽ 53%
Estimate the soil friction angle using
a. Eq. (1.83)
b. Eq. (1.84)
1.22 A consolidated-drained triaxial test on a normally consolidated clay yields the following results.
All around confining pressure, ␴3⬘ ⫽ 138 kN/m2
Added axial stress at failure, ⌬␴ ⫽ 276 kN/m2
Determine the shear-strength parameters.
1.23 The following are the results of two consolidated-drained triaxial tests on a clay.
Test I: ␴3⬘ ⫽ 82.8 kN/m2; ␴1⬘(failure) ⫽ 329.2 kN/m2
Test II: ␴3⬘ ⫽ 165.6 kN/m2; ␴1⬘(failure) ⫽ 558.6 kN/m2
Determine the shear-strength parameters—that is, c⬘ and ␾⬘.
1.24 A consolidated-undrained triaxial test was conducted on a saturated normally consolidated clay. The test results are
␴3 ⫽ 89.7 kN/m2
␴1(failure) ⫽ 220.8 kN/m2
Pore water pressure at failure ⫽ 37.95 kN/m2
Determine c, ␾, c⬘, and ␾⬘.
1.25 A normally consolidated clay soil has ␾ ⫽ 20° and ␾⬘ ⫽ 28°. If a consolidatedundrained test is conducted on this clay with an all around pressure of ␴3 ⫽
148.35 kN/m2, what would be the magnitude of the major principal stress, ␴1,
and the pore water pressure, u, at failure?
62 Chapter 1: Geotechnical Properties of Soil
References
AMER, A. M., and AWAD, A. A. (1974). “Permeability of Cohesionless Soils,” Journal of the
Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 100, No.
GT12, pp. 1309–1316.
AMERICAN SOCIETY FOR TESTING AND MATERIALS (2009). Annual Book of ASTM Standards,
Vol. 04.08, West Conshohocken, PA.
BJERRUM, L., and SIMONS, N. E. (1960). “Comparison of Shear Strength Characteristics of Normally
Consolidated Clay,” Proceedings, Research Conference on Shear Strength of Cohesive Soils,
ASCE, pp. 711–726.
CARRIER III, W. D. (2003). “Goodbye, Hazen; Hello, Kozeny-Carman,” Journal of Geotechnical and
Geoenvironmental Engineering, ASCE, Vol. 129, No. 11, pp. 1054–1056.
CASAGRANDE, A. (1936). “Determination of the Preconsolidation Load and Its Practical Significance,”
Proceedings, First International Conference on Soil Mechanics and Foundation Engineering,
Cambridge, MA, Vol. 3, pp. 60–64.
CHANDLER, R. J. (1988). “The in situ Measurement of the Undrained Shear Strength of Clays Using
the Field Vane,” STP 1014, Vane Shear Strength Testing in Soils: Field and Laboratory
Studies, American Society for Testing and Materials, pp. 13–44.
CHAPUIS, R. P. (2004). “Predicting the Saturated Hydraulic Conductivity of Sand and Gravel Using
Effective Diameter and Void Ratio,” Canadian Geotechnical Journal, Vol. 41, No. 5,
pp. 787–795.
DARCY, H. (1856). Les Fontaines Publiques de la Ville de Dijon, Paris.
DAS, B. M. (2009). Soil Mechanics Laboratory Manual, 7th ed., Oxford University Press, New York.
HANSBO, S. (1975). Jordmateriallära: 211, Stockholm, Awe/Gebers.
HIGHWAY RESEARCH BOARD (1945). Report of the Committee on Classification of Materials for
Subgrades and Granular Type Roads, Vol. 25, pp. 375– 388.
JAMILKOWSKI, M., LADD, C. C., GERMAINE, J. T., and LANCELLOTTA, R. (1985). “New Developments
in Field and Laboratory Testing of Soils,” Proceedings, XI International Conference on Soil
Mechanics and Foundations Engineering, San Francisco, Vol. 1, pp. 57–153.
KENNEY, T. C. (1959). “Discussion,” Journal of the Soil Mechanics and Foundations Division,
American Society of Civil Engineers, Vol. 85, No. SM3, pp. 67–69.
KULHAWY, F. H., and MAYNE, P. W. (1990). Manual of Estimating Soil Properties for Foundation
Design, Electric Power Research Institute, Palo Alto, California.
LADD, C. C., FOOTE, R., ISHIHARA, K., SCHLOSSER, F., and POULOS, H. G. (1977). “Stress Deformation and Strength Characteristics,” Proceedings, Ninth International Conference on Soil
Mechanics and Foundation Engineering, Tokyo, Vol. 2, pp. 421–494.
MESRI, G. (1989). “A Re-evaluation of su(mob) ⬇ 0.22␴p Using Laboratory Shear Tests,” Canadian
Geotechnical Journal, Vol. 26, No. 1, pp. 162–164.
MESRI, G., and OLSON, R. E. (1971). “Mechanism Controlling the Permeability of Clays,” Clay and
Clay Minerals, Vol. 19, pp. 151–158.
NAGARAJ, T. S., and MURTHY, B. R. S. (1985). “Prediction of the Preconsolidation Pressure and
Recompression Index of Soils,” Geotechnical Testing Journal, American Society for Testing
and Materials, Vol. 8, No. 4, pp. 199–202.
OLSON, R. E. (1977). “Consolidation Under Time-Dependent Loading,” Journal of Geotechnical
Engineering, ASCE, Vol. 103, No. GT1, pp. 55–60.
PARK, J. H., and KOUMOTO, T. (2004). “New Compression Index Equation,” Journal of Geotechnical
and Geoenvironmental Engineering, ASCE, Vol. 130, No. 2, pp. 223–226.
RENDON-HERRERO, O. (1980). “Universal Compression Index Equation,” Journal of the Geotechnical
Engineering Division, American Society of Civil Engineers, Vol. 106, No. GT11,
pp. 1178–1200.
SAMARASINGHE, A. M., HUANG, Y. H., and DRNEVICH, V. P. (1982). “Permeability and Consolidation
of Normally Consolidated Soils,” Journal of the Geotechnical Engineering Division, ASCE,
Vol. 108, No. GT6, pp. 835 –850.
References
63
SCHMERTMANN, J. H. (1953). “Undisturbed Consolidation Behavior of Clay,” Transactions,
American Society of Civil Engineers, Vol. 120, p. 1201.
SIVARAM, B., and SWAMEE, A. (1977), “A Computational Method for Consolidation Coefficient,”
Soils and Foundations, Vol. 17, No. 2, pp. 48–52.
SKEMPTON, A. W. (1944). “Notes on the Compressibility of Clays,” Quarterly Journal of Geological
Society, London, Vol. C, pp. 119–135.
SKEMPTON, A. W. (1953). “The Colloidal Activity of Clays,” Proceedings, 3rd International Conference
on Soil Mechanics and Foundation Engineering, London, Vol. 1, pp. 57–61.
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Institute of Civil Engineers, London, Vol. 7, pp. 305 –307.
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Under Axial Uplift and Compression Loading, REPORT EL-3771, Electric Power Research
Institute, Palo Alto, California.
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nichtbindiger Böden mit verschiedener Kornverteilung. Ph.D. Thesis, Technical University of
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Engineering Properties of Soils,” Canadian Geotechnical Journal, Vol. 15, No. 2, pp. 137–145.
2
2.1
Natural Soil Deposits and Subsoil
Exploration
Introduction
To design a foundation that will support a structure, an engineer must understand the
types of soil deposits that will support the foundation. Moreover, foundation engineers
must remember that soil at any site frequently is nonhomogeneous; that is, the soil profile may vary. Soil mechanics theories involve idealized conditions, so the application of
the theories to foundation engineering problems involves a judicious evaluation of site
conditions and soil parameters. To do this requires some knowledge of the geological
process by which the soil deposit at the site was formed, supplemented by subsurface
exploration. Good professional judgment constitutes an essential part of geotechnical
engineering—and it comes only with practice.
This chapter is divided into two parts. The first is a general overview of natural soil
deposits generally encountered, and the second describes the general principles of subsoil
exploration.
Natural Soil Deposits
2.2
Soil Origin
Most of the soils that cover the earth are formed by the weathering of various rocks. There are
two general types of weathering: (1) mechanical weathering and (2) chemical weathering.
Mechanical weathering is a process by which rocks are broken down into smaller
and smaller pieces by physical forces without any change in the chemical composition.
Changes in temperature result in expansion and contraction of rock due to gain and loss of
heat. Continuous expansion and contraction will result in the development of cracks in
rocks. Flakes and large fragments of rocks are split. Frost action is another source of
mechanical weathering of rocks. Water can enter the pores, cracks, and other openings in
the rock. When the temperature drops, the water freezes, thereby increasing the volume by
about 9%. This results in an outward pressure from inside the rock. Continuous freezing
and thawing will result in the breakup of a rock mass. Exfoliation is another mechanical
64
2.2 Soil Origin
65
weathering process by which rock plates are peeled off from large rocks by physical
forces. Mechanical weathering of rocks also takes place due to the action of running water,
glaciers, wind, ocean waves, and so forth.
Chemical weathering is a process of decomposition or mineral alteration in which
the original minerals are changed into something entirely different. For example, the
common minerals in igneous rocks are quartz, feldspars, and ferromagnesian minerals.
The decomposed products of these minerals due to chemical weathering are listed in
Table 2.1.
Table 2.1 Some Decomposed Products of Minerals in Igneous Rock
Mineral
Decomposed Product
Quartz
Quartz (sand grains)
Potassium feldspar (KAlSi3O8)
and Sodium feldspar (NaAlSi3O8)
Kaolinite (clay)
Bauxite
Illite (clay)
Silica
Calcium feldspar (CaAl2Si2O8)
Silica
Calcite
Biotite
Clay
Limonite
Hematite
Silica
Calcite
Olivine (Mg, Fe)2SiO4
Limonite
Serpentine
Hematite
Silica
Most rock weathering is a combination of mechanical and chemical weathering. Soil
produced by the weathering of rocks can be transported by physical processes to other
places. The resulting soil deposits are called transported soils. In contrast, some soils stay
where they were formed and cover the rock surface from which they derive. These soils
are referred to as residual soils.
Transported soils can be subdivided into five major categories based on the transporting agent:
1.
2.
3.
4.
5.
Gravity transported soil
Lacustrine (lake) deposits
Alluvial or fluvial soil deposited by running water
Glacial deposited by glaciers
Aeolian deposited by the wind
In addition to transported and residual soils, there are peats and organic soils, which derive
from the decomposition of organic materials.
66 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Residual Soil
Residual soils are found in areas where the rate of weathering is more than the rate at
which the weathered materials are carried away by transporting agents. The rate of weathering is higher in warm and humid regions compared to cooler and drier regions and,
depending on the climatic conditions, the effect of weathering may vary widely.
Residual soil deposits are common in the tropics, on islands such as the Hawaiian
Islands, and in the southeastern United States. The nature of a residual soil deposit will
generally depend on the parent rock. When hard rocks such as granite and gneiss
undergo weathering, most of the materials are likely to remain in place. These soil
deposits generally have a top layer of clayey or silty clay material, below which are silty
or sandy soil layers. These layers in turn are generally underlain by a partially weathered rock and then sound bedrock. The depth of the sound bedrock may vary widely,
even within a distance of a few meters. Figure 2.1 shows the boring log of a residual soil
deposit derived from the weathering of granite.
In contrast to hard rocks, there are some chemical rocks, such as limestone, that are
chiefly made up of calcite (CaCO3 ) mineral. Chalk and dolomite have large concentrations
of dolomite minerals 3Ca Mg(CO3 ) 24. These rocks have large amounts of soluble materials,
Fines (percent passing No. 200 sieve)
0
20
40
60
80 100
0
Light brown
silty clay
(Unified Soil
Classification—
MH)
Light brown
clayey silt
(Unified Soil
Classification—
MH)
Silty sand
(Unified Soil
Classification—
SC to SP)
1
2
3
Depth (m)
2.3
4
5
6
Partially
decomposed
granite
7
8
Bedrock
9
Figure 2.1 Boring log for a residual soil derived from granite
2.4 Gravity Transported Soil
67
some of which are removed by groundwater, leaving behind the insoluble fraction of the
rock. Residual soils that derive from chemical rocks do not possess a gradual transition zone
to the bedrock, as seen in Figure 2.1. The residual soils derived from the weathering of limestone-like rocks are mostly red in color. Although uniform in kind, the depth of weathering
may vary greatly. The residual soils immediately above the bedrock may be normally consolidated. Large foundations with heavy loads may be susceptible to large consolidation settlements on these soils.
2.4
Gravity Transported Soil
Residual soils on a natural slope can move downwards. Cruden and Varnes (1996) proposed a velocity scale for soil movement on a slope, which is summarized in Table 2.2.
When residual soils move down a natural slope very slowly, the process is usually referred
to as creep. When the downward movement of soil is sudden and rapid, it is called a landslide. The deposits formed by down-slope creep and landslides are colluvium.
Table 2.2 Velocity Scale for Soil Movement on a Slope
Description
Very slow
Slow
Moderate
Rapid
Velocity (mm/sec)
5 ⫻ 10⫺5 to 5 ⫻ 10⫺7
5 ⫻ 10⫺3 to 5 ⫻ 10⫺5
5 ⫻ 10⫺1 to 5 ⫻ 10⫺3
5 ⫻ 101 to 5 ⫻ 10⫺1
Colluvium is a heterogeneous mixture of soils and rock fragments ranging from
clay-sized particles to rocks having diameters of one meter or more. Mudflows are one type
of gravity-transported soil. Flows are downward movements of earth that resemble a viscous fluid (Figure 2.2) and come to rest in a more dense condition. The soil deposits
derived from past mudflows are highly heterogeneous in composition.
Mud flow
Figure 2.2 Mudflow
68 Chapter 2: Natural Soil Deposits and Subsoil Exploration
2.5
Alluvial Deposits
Alluvial soil deposits derive from the action of streams and rivers and can be divided into
two major categories: (1) braided-stream deposits and (2) deposits caused by the meandering belt of streams.
Deposits from Braided Streams
Braided streams are high-gradient, rapidly flowing streams that are highly erosive and carry
large amounts of sediment. Because of the high bed load, a minor change in the velocity of
flow will cause sediments to deposit. By this process, these streams may build up a complex
tangle of converging and diverging channels separated by sandbars and islands.
The deposits formed from braided streams are highly irregular in stratification and
have a wide range of grain sizes. Figure 2.3 shows a cross section of such a deposit. These
deposits share several characteristics:
1. The grain sizes usually range from gravel to silt. Clay-sized particles are generally
not found in deposits from braided streams.
2. Although grain size varies widely, the soil in a given pocket or lens is rather uniform.
3. At any given depth, the void ratio and unit weight may vary over a wide range
within a lateral distance of only a few meters. This variation can be observed during
soil exploration for the construction of a foundation for a structure. The standard
penetration resistance at a given depth obtained from various boreholes will be
highly irregular and variable.
Alluvial deposits are present in several parts of the western United States, such as
Southern California, Utah, and the basin and range sections of Nevada. Also, a large amount
of sediment originally derived from the Rocky Mountain range was carried eastward to
form the alluvial deposits of the Great Plains. On a smaller scale, this type of natural soil
deposit, left by braided streams, can be encountered locally.
Meander Belt Deposits
The term meander is derived from the Greek word maiandros, after the Maiandros (now
Menderes) River in Asia, famous for its winding course. Mature streams in a valley curve
back and forth. The valley floor in which a river meanders is referred to as the meander
Fine sand
Gravel
Silt
Coarse sand
Figure 2.3 Cross section of a braided-stream deposit
2.5 Alluvial Deposits
69
Erosion
Deposition
(point bar)
Deposition
(point bar)
River
Figure 2.4 Formation of point
bar deposits and oxbow lake in a
meandering stream
Oxbow lake
Erosion
belt. In a meandering river, the soil from the bank is continually eroded from the points
where it is concave in shape and is deposited at points where the bank is convex in shape,
as shown in Figure 2.4. These deposits are called point bar deposits, and they usually consist of sand and silt-size particles. Sometimes, during the process of erosion and deposition, the river abandons a meander and cuts a shorter path. The abandoned meander, when
filled with water, is called an oxbow lake. (See Figure 2.4.)
During floods, rivers overflow low-lying areas. The sand and silt-size particles carried
by the river are deposited along the banks to form ridges known as natural levees (Figure 2.5).
Levee deposit
Clay plug
Backswamp deposit
Lake
River
Figure 2.5 Levee and backswamp deposit
70 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Table 2.3 Properties of Deposits within the Mississippi Alluvial Valley
Environment
Soil texture
Natural
water
content
(%)
Liquid
limit
Plasticity
index
Natural levee
Clay (CL)
Silt (ML)
25–35
15–35
35–45
NP–35
15–25
NP–5
Point bar
Silt (ML) and
silty sand (SM)
25–45
30–55
10–25
Abandoned
channel
Clay (CL, CH)
30–95
30–100
10–65
Backswamps
Clay (CH)
25–70
40–115
25–100
Swamp
Organic clay
(OH)
100–265
135–300
100–165
(Note: NP—Nonplastic)
Finer soil particles consisting of silts and clays are carried by the water farther onto the floodplains. These particles settle at different rates to form what is referred to as backswamp
deposits (Figure 2.5), often highly plastic clays.
Table 2.3 gives some properties of soil deposits found in natural levees, point bars,
abandoned channels, backswamps and swamps within the alluvial Mississippi Valley
(Kolb and Shockley, 1959).
2.6
Lacustrine Deposits
Water from rivers and springs flows into lakes. In arid regions, streams carry large amounts
of suspended solids. Where the stream enters the lake, granular particles are deposited in
the area forming a delta. Some coarser particles and the finer particles (that is, silt and
clay) that are carried into the lake are deposited onto the lake bottom in alternate layers of
coarse-grained and fine-grained particles. The deltas formed in humid regions usually have
finer grained soil deposits compared to those in arid regions.
Varved clays are alternate layers of silt and silty clay with layer thicknesses rarely
exceeding about 13 mm. The silt and silty clay that constitute the layers were carried into
fresh water lakes by melt water at the end of the Ice Age. The hydraulic conductivity
(Section 1.10) of varved clays exhibits a high degree of anisotropy.
2.7
Glacial Deposits
During the Pleistocene Ice Age, glaciers covered large areas of the earth. The glaciers
advanced and retreated with time. During their advance, the glaciers carried large amounts of
sand, silt, clay, gravel, and boulders. Drift is a general term usually applied to the deposits laid
down by glaciers. The drifts can be broadly divided into two major categories: (a) unstratified
drifts and (b) stratified drifts. A brief description of each category follows.
2.8 Aeolian Soil Deposits
71
Unstratified Drifts
The unstratified drifts laid down by melting glaciers are referred to as till. The physical characteristics of till may vary from glacier to glacier. Till is called clay till
because of the presence of the large amount of clay-sized particles in it. In some areas,
tills constitute large amounts of boulders, and they are referred to as boulder till. The
range of grain sizes in a given till varies greatly. The amount of clay-sized fractions
present and the plasticity indices of tills also vary widely. During the field exploration
program, erratic values of standard penetration resistance (Section 2.13) also may be
expected.
The land forms that developed from the till deposits are called moraines. A terminal
moraine (Figure 2.6) is a ridge of till that marks the maximum limit of a glacier’s advance.
Recessional moraines are ridges of till developed behind the terminal moraine at varying
distances apart. They are the result of temporary stabilization of the glacier during the
recessional period. The till deposited by the glacier between the moraines is referred to as
ground moraine (Figure 2.6). Ground moraines constitute large areas of the central United
States and are called till plains.
Terminal moraine
Outwash
Ground moraine
Outwash
plain
Figure 2.6 Terminal moraine, ground moraine, and outwash plain
Stratified Drifts
The sand, silt, and gravel that are carried by the melting water from the front of a glacier
are called outwash. The melted water sorts out the particles by the grain size and forms
stratified deposits. In a pattern similar to that of braided-stream deposits, the melted water
also deposits the outwash, forming outwash plains (Figure 2.6), also called glaciofluvial
deposits.
2.8
Aeolian Soil Deposits
Wind is also a major transporting agent leading to the formation of soil deposits. When
large areas of sand lie exposed, wind can blow the sand away and redeposit it elsewhere.
Deposits of windblown sand generally take the shape of dunes (Figure 2.7). As dunes are
formed, the sand is blown over the crest by the wind. Beyond the crest, the sand particles
roll down the slope. The process tends to form a compact sand deposit on the windward
side, and a rather loose deposit on the leeward side, of the dune.
Dunes exist along the southern and eastern shores of Lake Michigan, the Atlantic
Coast, the southern coast of California, and at various places along the coasts of Oregon and
72 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Sand particle
Wind
direction
Figure 2.7 Sand dune
Washington. Sand dunes can also be found in the alluvial and rocky plains of the western
United States. Following are some of the typical properties of dune sand:
1. The grain-size distribution of the sand at any particular location is surprisingly uniform. This uniformity can be attributed to the sorting action of the wind.
2. The general grain size decreases with distance from the source, because the wind
carries the small particles farther than the large ones.
3. The relative density of sand deposited on the windward side of dunes may be as
high as 50 to 65%, decreasing to about 0 to 15% on the leeward side.
Figure 2.8 shows a sand dune from the Thar Desert, which is a large and arid region
located in the northwestern part of India that covers an area of about 200,000 square
kilometers.
Loess is an aeolian deposit consisting of silt and silt-sized particles. The grain-size
distribution of loess is rather uniform. The cohesion of loess is generally derived from a clay
Figure 2.8 A sand dune from the Thar Desert, India (Courtesy of A. S. Wayal, K. J. Somaiya
Polytechnic, Mumbai, India)
2.10 Some Local Terms for Soils
73
coating over the silt-sized particles, which contributes to a stable soil structure in an unsaturated state. The cohesion may also be the result of the precipitation of chemicals leached
by rainwater. Loess is a collapsing soil, because when the soil becomes saturated, it loses
its binding strength between particles. Special precautions need to be taken for the construction of foundations over loessial deposits. There are extensive deposits of loess in the
United States, mostly in the midwestern states of Iowa, Missouri, Illinois, and Nebraska and
for some distance along the Mississippi River in Tennessee and Mississippi.
Volcanic ash (with grain sizes between 0.25 to 4 mm) and volcanic dust (with grain
sizes less than 0.25 mm) may be classified as wind-transported soil. Volcanic ash is a
lightweight sand or sandy gravel. Decomposition of volcanic ash results in highly plastic and compressible clays.
2.9
Organic Soil
Organic soils are usually found in low-lying areas where the water table is near or above
the ground surface. The presence of a high water table helps in the growth of aquatic plants
that, when decomposed, form organic soil. This type of soil deposit is usually encountered
in coastal areas and in glaciated regions. Organic soils show the following characteristics:
1. Their natural moisture content may range from 200 to 300%.
2. They are highly compressible.
3. Laboratory tests have shown that, under loads, a large amount of settlement is
derived from secondary consolidation.
2.10
Some Local Terms for Soils
Soils are sometimes referred to by local terms. The following are a few of these terms with
a brief description of each.
1. Caliche: a Spanish word derived from the Latin word calix, meaning lime. It is
mostly found in the desert southwest of the United States. It is a mixture of sand,
silt, and gravel bonded together by calcareous deposits. The calcareous deposits are
brought to the surface by a net upward migration of water. The water evaporates in
the high local temperature. Because of the sparse rainfall, the carbonates are not
washed out of the top layer of soil.
2. Gumbo: a highly plastic, clayey soil.
3. Adobe: a highly plastic, clayey soil found in the southwestern United States.
4. Terra Rossa: residual soil deposits that are red in color and derive from limestone
and dolomite.
5. Muck: organic soil with a very high moisture content.
6. Muskeg: organic soil deposit.
7. Saprolite: residual soil deposit derived from mostly insoluble rock.
8. Loam: a mixture of soil grains of various sizes, such as sand, silt, and clay.
9. Laterite: characterized by the accumulation of iron oxide (Fe2 O3) and aluminum
oxide (Al2 O3) near the surface, and the leaching of silica. Lateritic soils in Central
America contain about 80 to 90% of clay and silt-size particles. In the United States,
lateritic soils can be found in the southeastern states, such as Alabama, Georgia, and
the Carolinas.
74 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Subsurface Exploration
2.11
Purpose of Subsurface Exploration
The process of identifying the layers of deposits that underlie a proposed structure and their
physical characteristics is generally referred to as subsurface exploration. The purpose of
subsurface exploration is to obtain information that will aid the geotechnical engineer in
1.
2.
3.
4.
Selecting the type and depth of foundation suitable for a given structure.
Evaluating the load-bearing capacity of the foundation.
Estimating the probable settlement of a structure.
Determining potential foundation problems (e.g., expansive soil, collapsible soil,
sanitary landfill, and so on).
5. Determining the location of the water table.
6. Predicting the lateral earth pressure for structures such as retaining walls, sheet pile
bulkheads, and braced cuts.
7. Establishing construction methods for changing subsoil conditions.
Subsurface exploration may also be necessary when additions and alterations to existing
structures are contemplated.
2.12
Subsurface Exploration Program
Subsurface exploration comprises several steps, including the collection of preliminary
information, reconnaissance, and site investigation.
Collection of Preliminary Information
This step involves obtaining information regarding the type of structure to be built and its
general use. For the construction of buildings, the approximate column loads and their
spacing and the local building-code and basement requirements should be known. The
construction of bridges requires determining the lengths of their spans and the loading on
piers and abutments.
A general idea of the topography and the type of soil to be encountered near and
around the proposed site can be obtained from the following sources:
1. United States Geological Survey maps.
2. State government geological survey maps.
3. United States Department of Agriculture’s Soil Conservation Service county soil
reports.
4. Agronomy maps published by the agriculture departments of various states.
5. Hydrological information published by the United States Corps of Engineers, including records of stream flow, information on high flood levels, tidal records, and so on.
6. Highway department soil manuals published by several states.
The information collected from these sources can be extremely helpful in planning a site
investigation. In some cases, substantial savings may be realized by anticipating problems
that may be encountered later in the exploration program.
2.12 Subsurface Exploration Program
75
Reconnaissance
The engineer should always make a visual inspection of the site to obtain information
about
1. The general topography of the site, the possible existence of drainage ditches, abandoned dumps of debris, and other materials present at the site. Also, evidence of
creep of slopes and deep, wide shrinkage cracks at regularly spaced intervals may be
indicative of expansive soils.
2. Soil stratification from deep cuts, such as those made for the construction of nearby
highways and railroads.
3. The type of vegetation at the site, which may indicate the nature of the soil. For
example, a mesquite cover in central Texas may indicate the existence of expansive
clays that can cause foundation problems.
4. High-water marks on nearby buildings and bridge abutments.
5. Groundwater levels, which can be determined by checking nearby wells.
6. The types of construction nearby and the existence of any cracks in walls or other
problems.
The nature of the stratification and physical properties of the soil nearby also can be
obtained from any available soil-exploration reports on existing structures.
Site Investigation
The site investigation phase of the exploration program consists of planning, making test
boreholes, and collecting soil samples at desired intervals for subsequent observation and
laboratory tests. The approximate required minimum depth of the borings should be predetermined. The depth can be changed during the drilling operation, depending on the subsoil
encountered. To determine the approximate minimum depth of boring, engineers may use
the rules established by the American Society of Civil Engineers (1972):
1. Determine the net increase in the effective stress, Ds r, under a foundation with
depth as shown in Figure 2.9. (The general equations for estimating increases in
stress are given in Chapter 5.)
2. Estimate the variation of the vertical effective stress, sor , with depth.
D
σ
σo
Figure 2.9 Determination of the minimum depth of boring
76 Chapter 2: Natural Soil Deposits and Subsoil Exploration
3. Determine the depth, D 5 D1, at which the effective stress increase Dsr is equal to
( 101 )q (q 5 estimated net stress on the foundation).
4. Determine the depth, D 5 D2, at which Dsr>sor 5 0.05.
5. Choose the smaller of the two depths, D1 and D2, just determined as the approximate minimum depth of boring required, unless bedrock is encountered.
If the preceding rules are used, the depths of boring for a building with a width of 30 m
will be approximately the following, according to Sowers and Sowers (1970):
No. of stories
Boring depth
1
2
3
4
5
3.5 m
6m
10 m
16 m
24 m
To determine the boring depth for hospitals and office buildings, Sowers and Sowers
(1970) also used the following rules.
•
For light steel or narrow concrete buildings,
Db
S0.7
5a
(2.1)
where
Db ⫽ depth of boring
S ⫽ number of stories
a 5 3 if Db is in meters
•
For heavy steel or wide concrete buildings,
Db
S0.7
5b
(2.2)
where
b5
6 if Db is in meters
20 if Db is in feet
When deep excavations are anticipated, the depth of boring should be at least 1.5 times the
depth of excavation.
Sometimes, subsoil conditions require that the foundation load be transmitted to
bedrock. The minimum depth of core boring into the bedrock is about 3 m. If the bedrock
is irregular or weathered, the core borings may have to be deeper.
There are no hard-and-fast rules for borehole spacing. Table 2.4 gives some general
guidelines. Spacing can be increased or decreased, depending on the condition of the subsoil. If various soil strata are more or less uniform and predictable, fewer boreholes are
needed than in nonhomogeneous soil strata.
2.13 Exploratory Borings in the Field
77
Table 2.4 Approximate Spacing of Boreholes
Type of project
Spacing
(m)
Multistory building
One-story industrial plants
Highways
Residential subdivision
Dams and dikes
10–30
20–60
250–500
250–500
40–80
The engineer should also take into account the ultimate cost of the structure when
making decisions regarding the extent of field exploration. The exploration cost generally
should be 0.1 to 0.5% of the cost of the structure. Soil borings can be made by several
methods, including auger boring, wash boring, percussion drilling, and rotary drilling.
2.13
Exploratory Borings in the Field
Auger boring is the simplest method of making exploratory boreholes. Figure 2.10 shows two
types of hand auger: the posthole auger and the helical auger. Hand augers cannot be used for
advancing holes to depths exceeding 3 to 5 m. However, they can be used for soil exploration
work on some highways and small structures. Portable power-driven helical augers (76 mm
to 305 mm in diameter) are available for making deeper boreholes. The soil samples obtained
from such borings are highly disturbed. In some noncohesive soils or soils having low cohesion, the walls of the boreholes will not stand unsupported. In such circumstances, a metal
pipe is used as a casing to prevent the soil from caving in.
(a)
(b)
Figure 2.10 Hand tools: (a) posthole auger;
(b) helical auger
78 Chapter 2: Natural Soil Deposits and Subsoil Exploration
When power is available, continuous-flight augers are probably the most common
method used for advancing a borehole. The power for drilling is delivered by truck- or
tractor-mounted drilling rigs. Boreholes up to about 60 to 70 m can easily be made by
this method. Continuous-flight augers are available in sections of about 1 to 2 m with
either a solid or hollow stem. Some of the commonly used solid-stem augers have outside diameters of 66.68 mm, 82.55 mm, 101.6 mm, and 114.3 mm. Common commercially available hollow-stem augers have dimensions of 63.5 mm ID and 158.75 mm
OD, 69.85 mm ID and 177.8 OD, 76.2 mm ID and 203.2 OD, and 82.55 mm ID and
228.6 mm OD.
The tip of the auger is attached to a cutter head (Figure 2.11). During the drilling
operation (Figure 2.12), section after section of auger can be added and the hole
extended downward. The flights of the augers bring the loose soil from the bottom of
the hole to the surface. The driller can detect changes in the type of soil by noting
changes in the speed and sound of drilling. When solid-stem augers are used, the auger
must be withdrawn at regular intervals to obtain soil samples and also to conduct other
operations such as standard penetration tests. Hollow-stem augers have a distinct
advantage over solid-stem augers in that they do not have to be removed frequently for
Figure 2.11 Carbide-tipped cutting head on auger flight (Courtesy of Braja M. Das,
Henderson, NV)
2.13 Exploratory Borings in the Field
79
Figure 2.12 Drilling with continuous-flight augers (Danny R. Anderson, PE of Professional
Service Industries, Inc, El Paso, Texas.)
sampling or other tests. As shown schematically in Figure 2.13, the outside of the hollowstem auger acts as a casing.
The hollow-stem auger system includes the following components:
Outer component:
Inner component:
(a) hollow auger sections, (b) hollow auger cap, and
(c) drive cap
(a) pilot assembly, (b) center rod column, and
(c) rod-to-cap adapter
The auger head contains replaceable carbide teeth. During drilling, if soil samples are to be
collected at a certain depth, the pilot assembly and the center rod are removed. The soil sampler is then inserted through the hollow stem of the auger column.
Wash boring is another method of advancing boreholes. In this method, a casing
about 2 to 3 m long is driven into the ground. The soil inside the casing is then removed
by means of a chopping bit attached to a drilling rod. Water is forced through the
drilling rod and exits at a very high velocity through the holes at the bottom of the
chopping bit (Figure 2.14). The water and the chopped soil particles rise in the drill
hole and overflow at the top of the casing through a T connection. The washwater is
80 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Rope
Drive cap
Rod-to-cap
adapter
Auger connector
Hollow-stem
auger section
Derrick
Pressure
water
Tub
Center rod
Casing
Drill rod
Pilot assembly
Auger connector
Chopping bit
Auger head
Center head
Replaceable
carbide
auger tooth
Figure 2.13 Hollow-stem auger components
(After ASTM, 2001) (ASTM D4700-91: Standard
Guide for Soil Sampling from the Vadose Zone.
Copyright ASTM INTERNATIONAL. Reprinted
with permission.)
Driving shoe
Water jet at
high velocity
Figure 2.14 Wash boring
collected in a container. The casing can be extended with additional pieces as the borehole progresses; however, that is not required if the borehole will stay open and not
cave in. Wash borings are rarely used now in the United States and other developed
countries.
Rotary drilling is a procedure by which rapidly rotating drilling bits attached to the
bottom of drilling rods cut and grind the soil and advance the borehole. There are several
types of drilling bit. Rotary drilling can be used in sand, clay, and rocks (unless they are badly
fissured). Water or drilling mud is forced down the drilling rods to the bits, and the return
flow forces the cuttings to the surface. Boreholes with diameters of 50 to 203 mm can easily
be made by this technique. The drilling mud is a slurry of water and bentonite. Generally, it
is used when the soil that is encountered is likely to cave in. When soil samples are needed,
the drilling rod is raised and the drilling bit is replaced by a sampler. With the environmental drilling applications, rotary drilling with air is becoming more common.
Percussion drilling is an alternative method of advancing a borehole, particularly
through hard soil and rock. A heavy drilling bit is raised and lowered to chop the hard soil.
The chopped soil particles are brought up by the circulation of water. Percussion drilling
may require casing.
2.15 Split-Spoon Sampling
2.14
81
Procedures for Sampling Soil
Two types of soil samples can be obtained during subsurface exploration: disturbed and
undisturbed. Disturbed, but representative, samples can generally be used for the following types of laboratory test:
1.
2.
3.
4.
5.
Grain-size analysis
Determination of liquid and plastic limits
Specific gravity of soil solids
Determination of organic content
Classification of soil
Disturbed soil samples, however, cannot be used for consolidation, hydraulic conductivity,
or shear strength tests. Undisturbed soil samples must be obtained for these types of laboratory tests. Sections 2.15 through 2.18 describe some procedures for obtaining soil samples during field exploration.
2.15
Split-Spoon Sampling
Split-spoon samplers can be used in the field to obtain soil samples that are generally disturbed, but still representative. A section of a standard split-spoon sampler is shown in
Figure 2.15a. The tool consists of a steel driving shoe, a steel tube that is split longitudinally
Water
port
Head
457.2 mm
(18 in.)
Pin
76.2 mm
(3 in.)
50.8
mm
34.93
mm
(2 in.) in.)
(1-3/8
Ball valve
Drilling
rod Coupling
Threads Driving
shoe
Split
barrel
(a)
(b)
Figure 2.15 (a) Standard split-spoon sampler; (b) spring core catcher
50.8 mm
(2 in.)
82 Chapter 2: Natural Soil Deposits and Subsoil Exploration
in half, and a coupling at the top. The coupling connects the sampler to the drill rod. The
standard split tube has an inside diameter of 34.93 mm and an outside diameter of 50.8 mm;
however, samplers having inside and outside diameters up to 63.5 mm and 76.2 mm, respectively, are also available. When a borehole is extended to a predetermined depth, the drill
tools are removed and the sampler is lowered to the bottom of the hole. The sampler is driven into the soil by hammer blows to the top of the drill rod. The standard weight of the hammer is 622.72 N, and for each blow, the hammer drops a distance of 0.762 m. The number
of blows required for a spoon penetration of three 152.4-mm intervals are recorded. The
number of blows required for the last two intervals are added to give the standard penetration number, N, at that depth. This number is generally referred to as the N value (American
Society for Testing and Materials, 2001, Designation D-1586-99). The sampler is then withdrawn, and the shoe and coupling are removed. Finally, the soil sample recovered from the
tube is placed in a glass bottle and transported to the laboratory. This field test is called the
standard penetration test (SPT). Figure 2.16a and b show a split-spoon sampler unassembled
before and after sampling.
The degree of disturbance for a soil sample is usually expressed as
AR (%) 5
D2o 2 D2i
D2i
(100)
(2.3)
where
AR 5 area ratio (ratio of disturbed area to total area of soil)
Do 5 outside diameter of the sampling tube
Di 5 inside diameter of the sampling tube
When the area ratio is 10% or less, the sample generally is considered to be undisturbed.
For a standard split-spoon sampler,
A R (%) 5
(50.8) 2 2 (34.93) 2
(34.93) 2
(100) 5 111.5%
Figure 2.16 (a) Unassembled split-spoon sampler; (b) after sampling (Courtesy of Professional Service Industries, Inc.
(PSI), Waukesha, Wisconsin)
2.15 Split-Spoon Sampling
83
Hence, these samples are highly disturbed. Split-spoon samples generally are taken at intervals of about 1.5 m. When the material encountered in the field is sand (particularly fine
sand below the water table), recovery of the sample by a split-spoon sampler may be difficult. In that case, a device such as a spring core catcher may have to be placed inside the
split spoon (Figure 2.15b).
At this juncture, it is important to point out that several factors contribute to the variation of the standard penetration number N at a given depth for similar soil profiles.
Among these factors are the SPT hammer efficiency, borehole diameter, sampling method,
and rod length (Skempton, 1986; Seed, et al., 1985). The SPT hammer energy efficiency
can be expressed as
Er (%) 5
actual hammer energy to the sampler
3 100
input energy
Theoretical input energy 5 Wh
(2.4)
(2.5)
where
W 5 weight of the hammer < 0.623 kN
h 5 height of drop < 0.76 mm
So,
Wh 5 (0.623) (0.76) 5 0.474 kN-m
In the field, the magnitude of Er can vary from 30 to 90%. The standard practice now in
the U.S. is to express the N-value to an average energy ratio of 60% (<N60 ). Thus, correcting for field procedures and on the basis of field observations, it appears reasonable
to standardize the field penetration number as a function of the input driving energy and
its dissipation around the sampler into the surrounding soil, or
N60 5
NhH hB hS hR
60
(2.6)
where
N60 5 standard penetration number, corrected for field conditions
N 5 measured penetration number
hH 5 hammer efficiency (%)
hB 5 correction for borehole diameter
hS 5 sampler correction
hR 5 correction for rod length
Variations of hH, hB, hS, and hR, based on recommendations by Seed et al. (1985)
and Skempton (1986), are summarized in Table 2.5.
Correlations for N60 in Cohesive Soil
Besides compelling the geotechnical engineer to obtain soil samples, standard penetration
tests provide several useful correlations. For example, the consistency of clay soils can be
estimated from the standard penetration number, N60. In order to achieve that, Szechy and
Vargi (1978) calculated the consistency index (CI) as
84 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Table 2.5 Variations of hH, hB, hS, and hR [Eq. (2.6)]
1. Variation of hH
2. Variation of hB
Country
Hammer type
Hammer release
hH (%)
Japan
Donut
Donut
Safety
Donut
Donut
Donut
Donut
Free fall
Rope and pulley
Rope and pulley
Rope and pulley
Rope and pulley
Free fall
Rope and pulley
78
67
60
45
45
60
50
United States
Argentina
China
hS
Standard sampler
With liner for dense sand and clay
With liner for loose sand
1.0
0.8
0.9
CI 5
hB
60 –120
150
200
1
1.05
1.15
4. Variation of hR
3. Variation of hS
Variable
Diameter
mm
Rod length
m
hR
.10
6 –10
4–6
0– 4
1.0
0.95
0.85
0.75
LL 2 w
LL 2 PL
(2.7)
where
w 5 natural moisture content
LL 5 liquid limit
PL 5 plastic limit
The approximate correlation between CI, N60, and the unconfined compression strength
(qu) is given in Table 2.6.
Hara, et al. (1971) also suggested the following correlation between the undrained
shear strength of clay (cu) and N60:
cu
5 0.29N 0.72
60
pa
(2.8)
where pa 5 atmospheric pressure (< 100 kN>m2; < 2000 lb>in2 ).
Table 2.6 Approximate Correlation between CI, N60, and qu
Standard penetration
number, N60
,2
2–8
8–15
15–30
.30
Consistency
Very soft
Soft to medium
Stiff
Very stiff
Hard
CI
Unconfined compression
strength, qu
(kN/m2)
,0.5
0.5–0.75
0.75–1.0
1.0–1.5
.1.5
,25
25–80
80–150
150–400
.400
2.15 Split-Spoon Sampling
85
The overconsolidation ratio, OCR, of a natural clay deposit can also be correlated
with the standard penetration number. On the basis of the regression analysis of 110 data
points, Mayne and Kemper (1988) obtained the relationship
OCR 5 0.193¢
N60 0.689
≤
sor
(2.9)
where sor 5 effective vertical stress in MN>m2 .
It is important to point out that any correlation between cu, OCR, and N60 is only
approximate.
Correction for N60 in Granular Soil
In granular soils, the value of N is affected by the effective overburden pressure, sor . For
that reason, the value of N60 obtained from field exploration under different effective overburden pressures should be changed to correspond to a standard value of sor . That is,
(N1 ) 60 5 CNN60
(2.10)
where
(N1 ) 60 5 value of N60 corrected to a standard value of sor 3100 kN>m2 A2000 lb>ft 2 B4
CN 5 correction factor
N60 5 value of N obtained from field exploration [Eq. (2.6)]
In the past, a number of empirical relations were proposed for CN. Some of the relationships are given next. The most commonly cited relationships are those of Liao and
Whitman (1986) and Skempton (1986).
In the following relationships for CN, note that sor is the effective overburden pressure and pa 5 atmospheric pressure A< 100 kN>m2 B
Liao and Whitman’s relationship (1986):
CN 5
1
0.5
C sor S
¢ ≤
pa
(2.11)
Skempton’s relationship (1986):
CN 5
2
11 ¢
CN 5
sor
≤
pa
3
21 ¢
sor
≤
pa
(for normally consolidated fine sand)
(2.12)
(for normally consolidated coarse sand)
(2.13)
86 Chapter 2: Natural Soil Deposits and Subsoil Exploration
CN 5
1.7
sor
0.7 1 ¢ ≤
pa
(for overconsolidated sand)
(2.14)
Seed et al.’s relationship (1975):
CN 5 1 21.25 log¢
sor
≤
pa
(2.15)
Peck et al.’s relationship (1974):
CN 5 0.77 log
20
£ sor §
¢ ≤
pa
¢for
sor
$ 0.25≤
pa
(2.16)
Bazaraa (1967):
CN 5
CN 5
4
sor
1 1 4a b
pa
afor
4
3.25 1 a
sor
b
pa
sor
# 0.75b
pa
afor
(2.17)
sor
. 0.75b
pa
(2.18)
Table 2.7 shows the comparison of CN derived using various relationships cited above. It
can be seen that the magnitude of the correction factor estimated by using any one of the
relationships is approximately the same, considering the uncertainties involved in conducting the standard penetration tests. Hence, it is recommended that Eq. (2.11) may be
used for all calculations.
Table 2.7 Variation of CN
CN
so9
pa
0.25
0.50
0.75
1.00
1.50
2.00
3.00
4.00
Eq. (2.11)
Eq. (2.12)
2.00
1.41
1.15
1.00
0.82
0.71
0.58
0.50
1.60
1.33
1.14
1.00
0.80
0.67
0.50
0.40
Eq. (2.13)
1.33
1.20
1.09
1.00
0.86
0.75
0.60
0.60
Eq. (2.14)
Eq. (2.15)
Eq. (2.16)
Eqs. (2.17)
and (2.18)
1.78
1.17
1.17
1.00
0.77
0.63
0.46
0.36
1.75
1.38
1.15
1.00
0.78
0.62
0.40
0.25
1.47
1.23
1.10
1.00
0.87
0.77
0.63
0.54
2.00
1.33
1.00
0.94
0.84
0.76
0.65
0.55
2.15 Split-Spoon Sampling
87
Correlation between N60 and Relative Density of Granular Soil
An approximate relationship between the corrected standard penetration number and the relative density of sand is given in Table 2.8. The values are approximate primarily because the
effective overburden pressure and the stress history of the soil significantly influence the N60
values of sand. Kulhawy and Mayne (1990) modified an empirical relationship for relative
density that was given by Marcuson and Bieganousky (1977), which can be expressed as
Dr (%) 5 12.2 1 0.75B222N60 1 2311 2 711OCR 2 779¢
0.5
sor
(2.19)
≤ 2 50C2u R
pa
where
Dr ⫽ relative density
␴o⬘ ⫽ effective overburden pressure
Cu ⫽ uniformity coefficient of sand
preconsolidation pressure, scr
OCR 5
effective overburden pressure, sor
pa 5 atmospheric pressure
Meyerhof (1957) developed a correlation between Dr and N60 as
N60 5 c17 1 24a
sor
b d D2r
pa
or
0.5
Dr 5 μ
N60
s9o
c17 1 24a b d
pa
∂
(2.20)
Equation (2.20) provides a reasonable estimate only for clean, medium fine sand.
Cubrinovski and Ishihara (1999) also proposed a correlation between N60 and the
relative density of sand (Dr ) that can be expressed as
N60 ¢0.23 1
Dr (%) 5 D
9
0.06 1.7
≤
D50
0.5
1
T (100)
s
£ or ≥
pa
Table 2.8 Relation between the Corrected (N1 )60
Values and the Relative Density in Sands
Standard penetration
number, (N1)60
Approximate relative
density, Dr (%)
0–5
5–10
10–30
30–50
0–5
5–30
30–60
60–95
(2.21)
88 Chapter 2: Natural Soil Deposits and Subsoil Exploration
where
pa 5 atmospheric pressure (< 100 kN>m2 )
D50 5 sieve size through which 50% of the soil will pass (mm)
Kulhawy and Mayne (1990) correlated the corrected standard penetration number
and the relative density of sand in the form
0.5
(N1 ) 60
Dr (%) 5 C
S
CpCACOCR
(100)
(2.22)
where
CP 5 grain-size correlations factor 5 60 1 25 logD50
(2.23)
t
b
100
COCR 5 correlation factor for overconsolidation 5 OCR0.18
D50 5 diameter through which 50% soil will pass through (mm)
t 5 age of soil since deposition (years)
OCR 5 overconsolidation ratio
CA 5 correlation factor for aging 5 1.2 1 0.05 loga
(2.24)
(2.25)
Correlation between Angle of Friction and Standard Penetration Number
The peak friction angle, fr, of granular soil has also been correlated with N60 or (N1 ) 60 by
several investigators. Some of these correlations are as follows:
1. Peck, Hanson, and Thornburn (1974) give a correlation between N60 and fr in a
graphical form, which can be approximated as (Wolff, 1989)
fr (deg) 5 27.1 1 0.3N60 2 0.000543N604 2
(2.26)
2. Schmertmann (1975) provided the correlation between N60, sor , and fr. Mathematically,
the correlation can be approximated as (Kulhawy and Mayne, 1990)
fr 5 tan21
£
N60
12.2 1 20.3 a
where
N60 5 field standard penetration number
sor 5 effective overburden pressure
pa 5 atmospheric pressure in the same unit as sor
fr 5 soil friction angle
0.34
sor §
b
pa
(2.27)
2.16 Sampling with a Scraper Bucket
89
3. Hatanaka and Uchida (1996) provided a simple correlation between fr and (N1 ) 60
that can be expressed as
fr 5 "20(N1 ) 60 1 20
(2.28)
The following qualifications should be noted when standard penetration resistance
values are used in the preceding correlations to estimate soil parameters:
1. The equations are approximate.
2. Because the soil is not homogeneous, the values of N60 obtained from a given borehole vary widely.
3. In soil deposits that contain large boulders and gravel, standard penetration numbers
may be erratic and unreliable.
Although approximate, with correct interpretation the standard penetration test provides a good evaluation of soil properties. The primary sources of error in standard penetration tests are inadequate cleaning of the borehole, careless measurement of the blow
count, eccentric hammer strikes on the drill rod, and inadequate maintenance of water head
in the borehole.
Correlation between Modulus of Elasticity and Standard
Penetration Number
The modulus of elasticity of granular soils (Es) is an important parameter in estimating the
elastic settlement of foundations. A first order estimation for Es was given by Kulhawy and
Mayne (1990) as
Es
5 aN60
pa
(2.29)
where
pa 5 atmospheric pressure (same unit as Es )
5 for sands with fines
a 5 • 10 for clean normally consolidated sand
15 for clean overconsolidated sand
2.16
Sampling with a Scraper Bucket
When the soil deposits are sand mixed with pebbles, obtaining samples by split spoon with
a spring core catcher may not be possible because the pebbles may prevent the springs
from closing. In such cases, a scraper bucket may be used to obtain disturbed representative samples (Figure 2.17). The scraper bucket has a driving point and can be attached to
a drilling rod. The sampler is driven down into the soil and rotated, and the scrapings from
the side fall into the bucket.
90 Chapter 2: Natural Soil Deposits and Subsoil Exploration
S
Drill rod
S
Driving point
Section
at S – S
Figure 2.17 Scraper bucket
2.17
Sampling with a Thin-Walled Tube
Thin-walled tubes are sometimes referred to as Shelby tubes. They are made of
seamless steel and are frequently used to obtain undisturbed clayey soils. The most
common thin-walled tube samplers have outside diameters of 50.8 mm and 76.2 mm.
The bottom end of the tube is sharpened. The tubes can be attached to drill rods
(Figure 2.18). The drill rod with the sampler attached is lowered to the bottom of the
borehole, and the sampler is pushed into the soil. The soil sample inside the tube
is then pulled out. The two ends are sealed, and the sampler is sent to the laboratory
for testing. Figure 2.19 shows the sequence of sampling with a thin-walled tube in
the field.
Samples obtained in this manner may be used for consolidation or shear tests. A
thin-walled tube with a 50.8-mm (2-in.) outside diameter has an inside diameter of about
47.63 mm (178 in.). The area ratio is
A R (%) 5
D2o 2 D2i
D2i
(100) 5
(50.8) 2 2 (47.63) 2
(47.63) 2
(100) 5 13.75%
Increasing the diameters of samples increases the cost of obtaining them.
Drill rod
Figure 2.18 Thin-walled tube
Thin-walled tube
2.17 Sampling with a Thin-Walled Tube
91
(a)
(b)
Figure 2.19 Sampling with a thin-walled tube: (a) tube being attached to drill rod; (b) tube sampler
pushed into soil (Courtesy of Khaled Sobhan, Florida Atlantic Univetsity, Boca Raton, Florida)
92 Chapter 2: Natural Soil Deposits and Subsoil Exploration
(c)
Figure 2.19 (continued) (c) recovery of soil sample (Courtesy of Khaled Sobhan, Florida
Atlantic University, Boca Raton, Florida)
2.18
Sampling with a Piston Sampler
When undisturbed soil samples are very soft or larger than 76.2 mm in diameter, they tend
to fall out of the sampler. Piston samplers are particularly useful under such conditions. There
are several types of piston sampler; however, the sampler proposed by Osterberg (1952) is
the most useful. (see Figures 2.20a and 2.20b). It consists of a thin-walled tube with a piston. Initially, the piston closes the end of the tube. The sampler is lowered to the bottom of
the borehole (Figure 2.20a), and the tube is pushed into the soil hydraulically, past the piston. Then the pressure is released through a hole in the piston rod (Figure 2.20b). To a large
extent, the presence of the piston prevents distortion in the sample by not letting the soil
squeeze into the sampling tube very fast and by not admitting excess soil. Consequently,
samples obtained in this manner are less disturbed than those obtained by Shelby tubes.
2.19
Observation of Water Tables
The presence of a water table near a foundation significantly affects the foundation’s loadbearing capacity and settlement, among other things. The water level will change seasonally. In many cases, establishing the highest and lowest possible levels of water during the
life of a project may become necessary.
2.19 Observation of Water Tables
93
Water (in)
Drill rod
Water
(out)
Vent
Piston
(a)
Sample
(b)
Figure 2.20 Piston sampler: (a) sampler at the bottom of borehole; (b) tube pushed into the soil
hydraulically
If water is encountered in a borehole during a field exploration, that fact should be
recorded. In soils with high hydraulic conductivity, the level of water in a borehole will
stabilize about 24 hours after completion of the boring. The depth of the water table can
then be recorded by lowering a chain or tape into the borehole.
In highly impermeable layers, the water level in a borehole may not stabilize for
several weeks. In such cases, if accurate water-level measurements are required, a
piezometer can be used. A piezometer basically consists of a porous stone or a perforated
pipe with a plastic standpipe attached to it. Figure 2.21 shows the general placement of a
piezometer in a borehole. This procedure will allow periodic checking until the water
level stabilizes.
94 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Protective
cover
Piezometer
water level
Groundwater
level
Standpipe
Bentonite
cement
grout
Bentonite
plug
Filter tip
Sand
2.20
Figure 2.21 Casagrande-type piezometer (Courtesy
of N. Sivakugan, James Cook University, Australia.)
Vane Shear Test
The vane shear test (ASTM D-2573) may be used during the drilling operation to determine the in situ undrained shear strength (cu ) of clay soils—particularly soft clays. The
vane shear apparatus consists of four blades on the end of a rod, as shown in Figure 2.22.
The height, H, of the vane is twice the diameter, D. The vane can be either rectangular
or tapered (see Figure 2.22). The dimensions of vanes used in the field are given in
Table 2.9. The vanes of the apparatus are pushed into the soil at the bottom of a borehole without disturbing the soil appreciably. Torque is applied at the top of the rod to
rotate the vanes at a standard rate of 0.1°>sec. This rotation will induce failure in a soil
of cylindrical shape surrounding the vanes. The maximum torque, T, applied to cause
failure is measured. Note that
T 5 f(cu, H, and D)
(2.30)
95
L = 10D
2.20 Vane Shear Test
H = 2D
45
D
D
Rectangular vane
Tapered vane
Figure 2.22 Geometry of field
vane (After ASTM, 2001)
(Annual Book of ASTM Standards,
Vol. 04.08. Copyright ASTM
INTERNATIONAL. Reprinted with
permission.)
or
cu 5
T
K
(2.31)
where
T is in N # m, cu is in kN>m2, and
K 5 a constant with a magnitude depending on the dimension and shape
of the vane
The constant
K5 ¢
p
D2H
D
≤
¢
≤ ¢1 1
≤
6
2
3H
10
(2.32a)
96 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Table 2.9 ASTM Recommended Dimensions of Field Vanesa (Annual Book of ASTM Standards,
Vol. 04.08. Copyright ASTM INTERNATIONAL. Reprinted with permission.)
Casing size
AX
BX
NX
101.6 mmb
Diameter, D
mm
Height, H
mm
Thickness of blade
mm
Diameter of rod
mm
38.1
50.8
63.5
92.1
76.2
101.6
127.0
184.1
1.6
1.6
3.2
3.2
12.7
12.7
12.7
12.7
a
The selection of a vane size is directly related to the consistency of the soil being tested; that is,
the softer the soil, the larger the vane diameter should be.
b
Inside diameter.
where
D 5 diameter of vane in cm
H 5 measured height of vane in cm
If H>D 5 2, Eq. (2.32a) yields
K 5 366 3 1028D3
c
(cm)
(2.32b)
In English units, if cu and T in Eq. (2.31) are expressed in lb>ft 2 and lb-ft, respectively, then
K5 ¢
p
D2H
D
≤¢
≤ ¢1 1
≤
1728
2
3H
(2.33a)
If H>D 5 2, Eq. (2.33a) yields
K 5 0.0021D3
c
(in.)
(2.33b)
Field vane shear tests are moderately rapid and economical and are used extensively in
field soil-exploration programs. The test gives good results in soft and medium-stiff clays and
gives excellent results in determining the properties of sensitive clays.
Sources of significant error in the field vane shear test are poor calibration of torque
measurement and damaged vanes. Other errors may be introduced if the rate of rotation of
the vane is not properly controlled.
For actual design purposes, the undrained shear strength values obtained from field
vane shear tests 3cu(VST) 4 are too high, and it is recommended that they be corrected according to the equation
cu(corrected) 5 lcu(VST)
(2.34)
where l 5 correction factor.
2.20 Vane Shear Test
97
Several correlations have been given previously for the correction factor l. The
most commonly used correlation for l is that given by Bjerrum (1972), which can be
expressed as
l 5 1.7 2 0.54 log 3PI(%)4
(2.35a)
Morris and Williams (1994) provided the following correlations:
l 5 1.18e 20.08(PI) 1 0.57 (for PI . 5)
l 5 7.01e
20.08(LL)
1 0.57 (where LL is in %)
(2.35b)
(2.35c)
The field vane shear strength can be correlated with the preconsolidation pressure
and the overconsolidation ratio of the clay. Using 343 data points, Mayne and Mitchell
(1988) derived the following empirical relationship for estimating the preconsolidation
pressure of a natural clay deposit:
scr 5 7.043cu(field) 4 0.83
(2.36)
Here,
scr 5 preconsolidation pressure (kN>m2 )
cu(field) 5 field vane shear strength (kN>m2 )
The overconsolidation ratio, OCR, also can be correlated to cu(field) according to the
equation
OCR 5 b
cu(field)
sor
(2.37)
where sor 5 effective overburden pressure.
The magnitudes of b developed by various investigators are given below.
•
Mayne and Mitchell (1988):
b 5 223PI(%)4 20.48
•
Hansbo (1957):
222
w(%)
(2.39)
1
0.08 1 0.0055(PI)
(2.40)
b5
•
(2.38)
Larsson (1980):
b5
98 Chapter 2: Natural Soil Deposits and Subsoil Exploration
2.21
Cone Penetration Test
The cone penetration test (CPT), originally known as the Dutch cone penetration test, is a versatile sounding method that can be used to determine the materials in a soil profile and estimate their engineering properties. The test is also called the static penetration test, and no
boreholes are necessary to perform it. In the original version, a 60° cone with a base area of
10 cm2 was pushed into the ground at a steady rate of about 20 mm>sec, and the resistance to
penetration (called the point resistance) was measured.
The cone penetrometers in use at present measure (a) the cone resistance (qc ) to penetration developed by the cone, which is equal to the vertical force applied to the cone,
divided by its horizontally projected area; and (b) the frictional resistance (fc ), which is the
resistance measured by a sleeve located above the cone with the local soil surrounding it.
The frictional resistance is equal to the vertical force applied to the sleeve, divided by its
surface area—actually, the sum of friction and adhesion.
Generally, two types of penetrometers are used to measure qc and fc:
1. Mechanical friction-cone penetrometer (Figure 2.23). The tip of this penetrometer is
connected to an inner set of rods. The tip is first advanced about 40 mm, giving the
35.7 mm
15 mm
15 mm
12.5 mm
30 mm dia.
47 mm
52.5 mm
45 mm
187 mm
11.5 mm
133.5
mm
20 mm dia.
35.7 mm
25 mm
387 mm
266 mm
69 mm
23 mm dia.
33.5 mm
32.5 mm dia.
146 mm
35.7 mm dia.
30 mm 35 mm
60
Collapsed
Extended
Figure 2.23 Mechanical friction-cone penetrometer (After ASTM, 2001) (Annual Book of
ASTM Standards, Vol. 04.08. Copyright ASTM INTERNATIONAL. Reprinted with permission.)
2.21 Cone Penetration Test
99
cone resistance. With further thrusting, the tip engages the friction sleeve. As the inner
rod advances, the rod force is equal to the sum of the vertical force on the cone and
sleeve. Subtracting the force on the cone gives the side resistance.
2. Electric friction-cone penetrometer (Figure 2.24). The tip of this penetrometer is
attached to a string of steel rods. The tip is pushed into the ground at the rate of
20 mm>sec. Wires from the transducers are threaded through the center of the rods
and continuously measure the cone and side resistances. Figure 2.25 shows a
photograph of an electric friction-cone penetrometer.
7
8
6
5
3
4
3
2
1
35.6 mm
1
2
3
4
5
6
7
8
Conical point (10 cm2)
Load cell
Strain gauges
Friction sleeve (150 cm2)
Adjustment ring
Waterproof bushing
Cable
Connection with rods
Figure 2.24 Electric friction-cone penetrometer (After ASTM, 2001) (Annual Book of ASTM
Standards, Vol. 04.08. Copyright ASTM INTERNATIONAL. Reprinted with permission.)
Figure 2.25 Photograph of an electric friction-cone penetrometer (Courtesy of Sanjeev Kumar,
Southern Illinois University, Carbondale, Illinois)
100 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Figure 2.26 shows the sequence of a cone penetration test in the field. A truckmounted CPT rig is shown in Figure 2.26a. A hydraulic ram located inside the truck
pushes the cone into the ground. Figure 2.26b shows the cone penetrometer in the
truck being put in the proper location. Figure 2.26c shows the progress of the CPT.
(a)
Figure 2.26 Cone penetration test in field:
(a) mounted CPT rig; (b) cone penetrometer being
set in proper location (Courtesy of Sanjeev Kumar,
Southern Illinois University, Carbondale, Illinois)
(a)
2.21 Cone Penetration Test
101
Figure 2.26 (continued) (c) test in progress (Courtesy of Sanjeev Kumar, Southern Illinois
University, Carbondale, Illinois)
Figure 2.27 shows the results of penetrometer test in a soil profile with friction measurement by an electric friction-cone penetrometer.
Several correlations that are useful in estimating the properties of soils encountered during an exploration program have been developed for the point resistance (qc )
and the friction ratio (Fr ) obtained from the cone penetration tests. The friction ratio is
defined as
Fr 5
fc
frictional resistance
5
qc
cone resistance
(2.41)
In a more recent study on several soils in Greece, Anagnostopoulos et al. (2003) expressed
Fr as
Fr (%) 5 1.45 2 1.36 logD50 (electric cone)
(2.42)
Fr (%) 5 0.7811 2 1.611 logD50 (mechanical cone)
(2.43)
and
where D50 5 size through which 50% of soil will pass through (mm).
The D50 for soils based on which Eqs. (2.42) and (2.43) have been developed ranged
from 0.001 mm to about 10 mm.
As in the case of standard penetration tests, several correlations have been developed
between qc and other soil properties. Some of these correlations are presented next.
102 Chapter 2: Natural Soil Deposits and Subsoil Exploration
qc (kN/m2)
5,000
0
10,000
0
0
2
2
4
4
6
Depth (m)
Depth (m)
0
fc (kN/m2)
200
400
6
8
8
10
10
12
12
Figure 2.27 Cone penetrometer test with friction measurement
Correlation between Relative Density (Dr ) and qc for Sand
Lancellotta (1983) and Jamiolkowski et al. (1985) showed that the relative density of normally consolidated sand, Dr, and qc can be correlated according to the formula (Figure 2.28).
Dr (%) 5 A 1 B log 10 ¢
qc
"sor
≤
(2.44)
The preceding relationship can be rewritten as (Kulhawy and Mayne, 1990)
Dr (%) 5 68Clog £
qc
"pa ? s0r
≥ 2 1S
(2.45)
2.21 Cone Penetration Test
103
95
85
Dr = –98 + 66 log10
qc
(σ 0 )0.5
75
Dr (%)
65
2σ
55
qc and σ 0 in ton (metric)/m2
2σ
45
Ticino sand
Ottawa sand
Edgar sand
35
Hokksund sand
25
Hilton mine sand
15
1000
100
qc
σ 0
0.5
Figure 2.28 Relationship between Dr and qc (Based on Lancellotta, 1983,
and Jamiolski et al., 1985)
where
pa 5 atmospheric pressure (< 100 kN>m2 )
sor 5 vertical effective stress
Baldi et al. (1982), and Robertson and Campanella (1983) recommended the empirical relationship shown in Figure 2.29 between vertical effective stress (sor ), relative density
(Dr ), and qc for normally consolidated sand.
Kulhawy and Mayne (1990) proposed the following relationship to correlate Dr, qc,
and the vertical effective stress sor :
Dr 5
qc
pa
1
R≥
¥
305QcOCR1.8
sor 0.5
¢ ≤
ï
pa
B
In this equation,
OCR 5 overconsolidation ratio
pa 5 atmospheric pressure
Qc 5 compressibility factor
(2.46)
104 Chapter 2: Natural Soil Deposits and Subsoil Exploration
0
Cone point resistance, qc (MN/m2)
10
20
30
40
50
0
Vertical effective stress, σo (kN/m2)
100
200
300
400
500
Dr = 40% 50% 60% 70% 80% 90%
Figure 2.29 Variation of qc, sor , and
Dr for normally consolidated quartz sand
(Based on Baldi et al., 1982, and
Robertson and Campanella, 1983)
The recommended values of Qc are as follows:
Highly compressible sand 5 0.91
Moderately compressible sand 5 1.0
Low compressible sand 5 1.09
Correlation between qc and Drained Friction Angle (f9) for Sand
On the basis of experimental results, Robertson and Campanella (1983) suggested the variation of Dr, sor , and fr for normally consolidated quartz sand. This relationship can be
expressed as (Kulhawy and Mayne, 1990)
fr 5 tan21 B0.1 1 0.38 log¢
qc
≤R
sor
(2.47)
Based on the cone penetration tests on the soils in the Venice Lagoon (Italy), Ricceri et
al. (2002) proposed a similar relationship for soil with classifications of ML and SP-SM as
fr 5 tan21 B0.38 1 0.27 log¢
qc
≤R
sor
(2.48)
In a more recent study, Lee et al. (2004) developed a correlation between fr, qc , and the horizontal effective stress (shr ) in the form
fr 5 15.575¢
qc 0.1714
≤
shr
(2.49)
2.21 Cone Penetration Test
Clay
1000
Clayey silt & Sandy silt
silty clay
and silt
Silty Sand
105
Sand
900
qc (kN/m2)
N60
700
Ratio,
800
400
600
500
Range of results of
Robertson & Campanella (1983)
300
200
Average of Robertson &
Campanella (1983)
100
0
0.001
0.01
0.1
Mean grain size, D50 (mm)
1.0
Figure 2.30 General range of variation of qc>N60 for various
types of soil
Correlation between qc and N60
Figure 2.30 shows a plot of qc (kN> m2)> N60 (N60 5 standard penetration resistance)
against the mean grain size (D50 in mm) for various types of soil. This was developed from
field test results by Robertson and Campanella (1983).
Anagnostopoulos et al. (2003) provided a similar relationship correlating qc, N60 ,
and D50. Or
¢
qc
≤
pa
N60
5 7.6429D0.26
50
(2.50)
where pa 5 atmospheric pressure (same unit as qc).
Correlations of Soil Types
Robertson and Campanella (1983) provided the correlations shown in Figure 2.31 between
qc and the friction ratio [Eq. (2.41)] to identify various types of soil encountered in the field.
Correlations for Undrained Shear Strength (cu), Preconsolidation
Pressure (scr ), and Overconsolidation Ratio (OCR) for Clays
The undrained shear strength, cu, can be expressed as
cu 5
qc 2 so
NK
(2.51)
106 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Cone point resistance, qc (MN/m2)
40
Sands
20
Silty
sands
10
8
6
4
Sandy
Clayey
silts
silts
and
and
silts
silty
clays
Clays
2
1
0.8
0.6
0.4
Peat
0.2
0.1
0
1
2
3
4
5
Friction ratio, Fr (%)
6
Figure 2.31 Robertson and Campanella’s
correlation (1983) between qc, Fr, and the
type of soil (Robertson and Campanella, 1983)
where
so 5 total vertical stress
NK 5 bearing capacity factor
The bearing capacity factor, NK, may vary from 11 to 19 for normally consolidated clays and
may approach 25 for overconsolidated clay. According to Mayne and Kemper (1988)
NK 5 15 (for electric cone)
and
NK 5 20 (for mechanical cone)
Based on tests in Greece, Anagnostopoulos et al. (2003) determined
NK 5 17.2 (for electric cone)
and
NK 5 18.9 (for mechanical cone)
These field tests also showed that
cu 5
fc
(for mechanical cones)
1.26
(2.52)
and
cu 5 fc (for electrical cones)
(2.53)
Mayne and Kemper (1988) provided correlations for preconsolidation pressure (scr)
and overconsolidation ratio (OCR) as
scr 5 0.243(qc ) 0.96
c
c
2
MN>m
MN>m2
(2.54)
2.22 Pressuremeter Test (PMT)
107
and
OCR 5 0.37¢
qc 2 so 1.01
≤
sor
(2.55)
where so and sor 5 total and effective stress, respectively.
2.22
Pressuremeter Test (PMT)
The pressuremeter test is an in situ test conducted in a borehole. It was originally developed by
Menard (1956) to measure the strength and deformability of soil. It has also been adopted by
ASTM as Test Designation 4719. The Menard-type PMT consists essentially of a probe with
three cells. The top and bottom ones are guard cells and the middle one is the measuring cell, as
shown schematically in Figure 2.32a. The test is conducted in a prebored hole with a diameter
that is between 1.03 and 1.2 times the nominal diameter of the probe. The probe that is most
commonly used has a diameter of 58 mm and a length of 420 mm. The probe cells can be
expanded by either liquid or gas. The guard cells are expanded to reduce the end-condition effect
on the measuring cell, which has a volume (Vo ) of 535 cm3. Following are the dimensions for
the probe diameter and the diameter of the borehole, as recommended by ASTM:
Probe
diameter
(mm)
Nominal (mm)
Maximum (mm)
44
58
74
45
60
76
53
70
89
Gas/water
line
Borehole diameter
pl
Pressure, p
Zone I
Zone II
Zone III
pf
Guard
cell
p
Measuring
cell
Guard
cell
po
v
Vo
(a)
Vo υo Vo υm Vo υf
(b)
2(Vo υo)
Figure 2.32 (a) Pressuremeter; (b) plot of pressure versus total cavity volume
Total
cavity
volume,
V
108 Chapter 2: Natural Soil Deposits and Subsoil Exploration
In order to conduct a test, the measuring cell volume, Vo, is measured and the probe
is inserted into the borehole. Pressure is applied in increments and the new volume of the
cell is measured. The process is continued until the soil fails or until the pressure limit of
the device is reached. The soil is considered to have failed when the total volume of the
expanded cavity (V) is about twice the volume of the original cavity. After the completion
of the test, the probe is deflated and advanced for testing at another depth.
The results of the pressuremeter test are expressed in the graphical form of pressure
versus volume, as shown in Figure 2.32b. In the figure, Zone I represents the reloading portion during which the soil around the borehole is pushed back into the initial state (i.e., the
state it was in before drilling). The pressure po represents the in situ total horizontal stress.
Zone II represents a pseudoelastic zone in which the cell volume versus cell pressure is
practically linear. The pressure pf represents the creep, or yield, pressure. The zone marked
III is the plastic zone. The pressure pl represents the limit pressure. Figure 2.33 shows some
photographs for a pressuremeter test in the field.
The pressuremeter modulus, Ep, of the soil is determined with the use of the theory
of expansion of an infinitely thick cylinder. Thus,
Ep 5 2(1 1 ms ) (Vo 1 vm ) ¢
Dp
≤
Dv
(2.56)
where
vm 5
vo 1 vf
2
Dp 5 pf 2 po
Dv 5 vf 2 vo
ms 5 Poisson’s ratio (which may be assumed to be 0.33)
The limit pressure pl is usually obtained by extrapolation and not by direct measurement.
In order to overcome the difficulty of preparing the borehole to the proper size, selfboring pressuremeters (SBPMTs) have also been developed. The details concerning
SBPMTs can be found in the work of Baguelin et al. (1978).
Correlations between various soil parameters and the results obtained from the pressuremeter tests have been developed by various investigators. Kulhawy and Mayne (1990)
proposed that, for clays,
scr 5 0.45pl
(2.57)
where scr 5 preconsolidation pressure.
On the basis of the cavity expansion theory, Baguelin et al. (1978) proposed that
cu 5
(pl 2 po )
Np
(2.58)
2.22 Pressuremeter Test (PMT)
(a)
(c)
(b)
(d)
109
Figure 2.33 Pressuremeter test in the field: (a) the pressuremeter probe; (b) drilling the bore hole
by wet rotary method; (c) pressuremeter control unit with probe in the background; (d) getting
ready to insert the pressuremeter probe into the bore hole (Courtesy of Jean-Louis Briaud, Texas
A&M University, College Station, Texas)
110 Chapter 2: Natural Soil Deposits and Subsoil Exploration
where
cu 5 undrained shear strength of a clay
Ep
Np 5 1 1 ln ¢
≤
3cu
Typical values of Np vary between 5 and 12, with an average of about 8.5. Ohya et al.
(1982) (see also Kulhawy and Mayne, 1990) correlated Ep with field standard penetration
numbers (N60 ) for sand and clay as follows:
Clay: Ep (kN>m2 ) 5 1930N 0.63
60
2
Sand: Ep (kN>m ) 5
2.23
908N 0.66
60
(2.59)
(2.60)
Dilatometer Test
The use of the flat-plate dilatometer test (DMT) is relatively recent (Marchetti, 1980;
Schmertmann, 1986). The equipment essentially consists of a flat plate measuring 220 mm
(length) 3 95 mm (width) 3 14 mm (thickness). A thin, flat, circular, expandable steel
membrane having a diameter of 60 mm is located flush at the center on one side of the plate
(Figure 2.34a). Figure 2.35 shows two flat-plate dilatometers with other instruments for
conducting a test in the field. The dilatometer probe is inserted into the ground with a cone
penetrometer testing rig (Figure 2.34b). Gas and electric lines extend from the surface control
box, through the penetrometer rod, and into the blade. At the required depth, high-pressure
nitrogen gas is used to inflate the membrane. Two pressure readings are taken:
1. The pressure A required to “lift off” the membrane.
2. The pressure B at which the membrane expands 1.1 mm into the surrounding soil
60
mm
95 mm
(a)
(b)
Figure 2.34 (a) Schematic diagram of a
flat-plate dilatometer; (b) dilatometer probe
inserted into ground
2.23 Dilatometer Test
111
Figure 2.35 Dilatometer and other equipment (Courtesy of N. Sivakugan, James Cook University,
Australia)
The A and B readings are corrected as follows (Schmertmann, 1986):
Contact stress, po 5 1.05(A 1 DA 2 Zm ) 2 0.05(B 2 DB 2 Zm )
Expansion stress, p1 5 B 2 Zm 2 DB
(2.61)
(2.62)
where
DA 5 vacuum pressure required to keep the membrane in contact with its seating
DB 5 air pressure required inside the membrane to deflect it outward to a center expansion of 1.1 mm
Zm 5 gauge pressure deviation from zero when vented to atmospheric pressure
The test is normally conducted at depths 200 to 300 mm apart. The result of a given test is
used to determine three parameters:
1. Material index, ID 5
p1 2 po
po 2 uo
2. Horizontal stress index, KD 5
po 2 uo
sor
3. Dilatometer modulus, ED (kN>m2 ) 5 34.7(p1 kN>m2 2 po kN>m2 )
112 Chapter 2: Natural Soil Deposits and Subsoil Exploration
where
uo 5 pore water pressure
sor 5 in situ vertical effective stress
Figure 2.36 shows the results of a dilatometer test conducted in Bangkok soft clay
and reported by Shibuya and Hanh (2001). Based on his initial tests, Marchetti (1980) provided the following correlations.
KD 0.47
≤ 2 0.6
1.5
(2.63)
OCR 5 (0.5KD ) 1.56
(2.64)
Ko 5 ¢
cu
5 0.22
sor
0
pO , p1 (kN/m2)
300
600 0
ID
0.3
(for normally consolidated clay)
0.6
0
KD
3
6
0
(2.65)
ED (kN/m2)
2,000
4,000 5,000
0
2
4
Depth (m)
6
8
p1
10
pO
12
14
Figure 2.36 A dilatometer test result conducted on soft Bangkok clay (Redrawn from Shibuya
and Hanh, 2001)
2.24 Coring of Rocks
¢
113
cu
cu
≤ 5 ¢ ≤ (0.5KD ) 1.25
sor OC
sor NC
(2.66)
Es 5 (1 2 m2s )ED
(2.67)
where
Ko 5 coefficient of at-rest earth pressure
OCR 5 overconsolidation ratio
OC 5 overconsolidated soil
NC 5 normally consolidated soil
Es 5 modulus of elasticity
Other relevant correlations using the results of dilatometer tests are as follows:
•
For undrained cohesion in clay (Kamei and Iwasaki, 1995):
cu 5 0.35 s0r (0.47KD ) 1.14
•
(2.68)
For soil friction angle (ML and SP-SM soils) (Ricceri et al., 2002):
fr 5 31 1
KD
0.236 1 0.066KD
fult
r 5 28 1 14.6 logKD 2 2.1(logKD ) 2
(2.69a)
(2.69b)
Schmertmann (1986) also provided a correlation between the material index (ID )
and the dilatometer modulus (ED ) for a determination of the nature of the soil and its unit
weight (g). This relationship is shown in Figure 2.37.
2.24
Coring of Rocks
When a rock layer is encountered during a drilling operation, rock coring may be necessary. To core rocks, a core barrel is attached to a drilling rod. A coring bit is attached to
the bottom of the barrel (Fig. 2.38). The cutting elements may be diamond, tungsten, carbide, and so on. Table 2.10 summarizes the various types of core barrel and their sizes, as
well as the compatible drill rods commonly used for exploring foundations. The coring is
advanced by rotary drilling. Water is circulated through the drilling rod during coring, and
the cutting is washed out.
114 Chapter 2: Natural Soil Deposits and Subsoil Exploration
200
Clay
Silt
Silty
Clayey
Sand
Sandy
Silty
100
Dilatometer modulus, ED (MN/m2)
gid
Ri 0)
0
(2.
e
ns
de
ry 0)
e
V 2.1
(
50
5
ty
idi
rig
w
Lo 1.80)
(
ty
nsi
de
m
diu .80)
e
M
(1
cy
ten
sis
on )
c
gh 1.90
(
Hi
cy
ten
sis
n
o
c
)
m
diu (1.80
Me
cy
ten
sis
n
o
wc
0)*
Lo (1.7
10
ity
id
rig
um 0)
i
d
Me (1.9
nse
De 5)
9
(1.
rd
Ha 5)
0
(2.
20
id
rig
ry
Ve .15)
(2
ose
Lo 0)
7
(1.
ity
ens
w d 0)
o
L
7
(1.
le
sib
res
p
)
m
Co (1.60
ft
So )*
60
(1.
2
1.2
1.0
0.35
0.6
Mud/Peat
(1.50)
0.5
10
0.2
0.5
0.9
3.3
1.2
1.8
(␥) – Approximate soil unit weight in
t/m3 shown in parentheses
* – If PI > 50, then ␥ in these regions is
overestimated by about 0.10 t/m3
1
2
Material index, ID
5
10
Figure 2.37 Chart for determination
of soil description and unit weight
(After Schmertmann, 1986)
(Note: 1 t>m3 5 9.81 kN>m3)
(Schmertmann, J. H. (1986). “Suggested
method for performing the flat dilatometer
test,” Geotechnical Testing Journal,
ASTM, Vol. 9, No. 2, pp. 93-101, Fig. 2.
Copyright ASTM INTERNATIONAL.
Reprinted with permission.)
Table 2.10 Standard Size and Designation of Casing, Core Barrel, and Compatible Drill Rod
Casing and
core barrel
designation
Outside
diameter of
core barrel bit
(mm)
EX
AX
BX
NX
36.51
47.63
58.74
74.61
Drill rod
designation
Outside
diameter of
drill rod
(mm)
Diameter of
borehole
(mm)
Diameter of
core sample
(mm)
E
A
B
N
33.34
41.28
47.63
60.33
38.1
50.8
63.5
76.2
22.23
28.58
41.28
53.98
Two types of core barrel are available: the single-tube core barrel (Figure 2.38a)
and the double-tube core barrel (Figure 2.38b). Rock cores obtained by single-tube
core barrels can be highly disturbed and fractured because of torsion. Rock cores
smaller than the BX size tend to fracture during the coring process. Figure 2.39 shows
2.24 Coring of Rocks
Drill rod
Drill rod
Inner
barrel
Core
barrel
Rock
Rock
Rock
core
115
Outer
barrel
Rock
Rock
core
Core
lifter
Coring
bit
Core
lifter
Coring
bit
(a)
(b)
Figure 2.38 Rock coring: (a) singletube core barrel; (b) double-tube core
barrel
Figure 2.39 Diamond coring bit
(Courtesy of Braja M. Das,
Henderson, NV)
116 Chapter 2: Natural Soil Deposits and Subsoil Exploration
(a)
(b)
Figure 2.40 Diamond coring bit attached to a double-tube core barrel: (a) end view;
(b) side view (Courtesy of Professional Service Industries, Inc. (PSI), Waukesha, Wisconsin)
the photograph of a diamond coring bit. Figure 2.40 shows the end and side views of a
diamond coring bit attached to a double-tube core barrel.
When the core samples are recovered, the depth of recovery should be properly
recorded for further evaluation in the laboratory. Based on the length of the rock core
2.25 Preparation of Boring Logs
117
recovered from each run, the following quantities may be calculated for a general evaluation of the rock quality encountered:
Recovery ratio 5
length of core recovered
theoretical length of rock cored
(2.70)
Rock quality designation (RQD)
5
S length of recovered pieces equal to or larger than 101.6 mm
theoretical length of rock cored
(2.71)
A recovery ratio of unity indicates the presence of intact rock; for highly fractured
rocks, the recovery ratio may be 0.5 or smaller. Table 2.11 presents the general relationship (Deere, 1963) between the RQD and the in situ rock quality.
Table 2.11 Relation between in situ
Rock Quality and RQD
2.25
RQD
Rock quality
0– 0.25
0.25– 0.5
0.5–0.75
0.75– 0.9
0.9–1
Very poor
Poor
Fair
Good
Excellent
Preparation of Boring Logs
The detailed information gathered from each borehole is presented in a graphical form
called the boring log. As a borehole is advanced downward, the driller generally should
record the following information in a standard log:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Name and address of the drilling company
Driller’s name
Job description and number
Number, type, and location of boring
Date of boring
Subsurface stratification, which can be obtained by visual observation of the soil
brought out by auger, split-spoon sampler, and thin-walled Shelby tube sampler
Elevation of water table and date observed, use of casing and mud losses, and so on
Standard penetration resistance and the depth of SPT
Number, type, and depth of soil sample collected
In case of rock coring, type of core barrel used and, for each run, the actual length
of coring, length of core recovery, and RQD
This information should never be left to memory, because doing so often results in erroneous boring logs.
118 Chapter 2: Natural Soil Deposits and Subsoil Exploration
Boring Log
Name of the Project
Two-story apartment building
Location Johnson & Olive St. Date of Boring March 2, 2005
60.8 m
Boring No. 3 Type of Hollow-stem auger Ground
Boring
Elevation
Soil
wn
Soil
Depth sample N
Comments
60
(%)
description
(m) type and
number
Light brown clay (fill)
1
Silty sand (SM)
G.W.T.
3.5 m
Light gray clayey
silt (ML)
2
3
9
8.2
SS-2
12
17.6
LL 38
PI 11
20.4
LL 36
qu 112 kN/m2
4
5
6
Sand with some
gravel (SP)
End of boring @ 8 m
SS-1
ST-1
SS-3
11
20.6
7
SS-4
27
8
N60 standard penetration number
wn natural moisture content
LL liquid limit; PI plasticity index
qu unconfined compression strength
SS split-spoon sample; ST Shelby tube sample
9
Groundwater table
observed after one
week of drilling
Figure 2.41 A typical boring log
After completion of the necessary laboratory tests, the geotechnical engineer prepares a finished log that includes notes from the driller’s field log and the results of tests
conducted in the laboratory. Figure 2.41 shows a typical boring log. These logs have to be
attached to the final soil-exploration report submitted to the client. The figure also lists the
classifications of the soils in the left-hand column, along with the description of each soil
(based on the Unified Soil Classification System).
2.26
Geophysical Exploration
Several types of geophysical exploration techniques permit a rapid evaluation of subsoil
characteristics. These methods also allow rapid coverage of large areas and are less expensive than conventional exploration by drilling. However, in many cases, definitive interpretation of the results is difficult. For that reason, such techniques should be used for
preliminary work only. Here, we discuss three types of geophysical exploration technique:
the seismic refraction survey, cross-hole seismic survey, and resistivity survey.
Seismic Refraction Survey
Seismic refraction surveys are useful in obtaining preliminary information about the
thickness of the layering of various soils and the depth to rock or hard soil at a site.
2.26 Geophysical Exploration
119
Refraction surveys are conducted by impacting the surface, such as at point A in
Figure 2.42a, and observing the first arrival of the disturbance (stress waves) at several other points (e.g., B, C, D, c ). The impact can be created by a hammer blow or
by a small explosive charge. The first arrival of disturbance waves at various points
can be recorded by geophones.
The impact on the ground surface creates two types of stress wave: P waves (or
plane waves) and S waves (or shear waves). P waves travel faster than S waves; hence, the
first arrival of disturbance waves will be related to the velocities of the P waves in various
layers. The velocity of P waves in a medium is
v5
Es
g
ã ¢g≤
(1 2 ms )
(1 2 2ms ) (1 1 ms )
where
Es 5 modulus of elasticity of the medium
g 5 unit weight of the medium
g 5 acceleration due to gravity
ms 5 Poisson’s ratio
x
(x1) B
A v1
Layer I
v1
(x2) C
v1
(x3) D
v1
v1
Z1
Velocity
v1
v2
v2
Layer II
v2
v3
Layer III
Velocity
v3
Time of first arrival
(a)
d
c
Ti2
b
Ti1
xc
a
Distance, x
(b)
Figure 2.42 Seismic refraction survey
Z2 Velocity
v2
(2.72)
120 Chapter 2: Natural Soil Deposits and Subsoil Exploration
To determine the velocity v of P waves in various layers and the thicknesses of those
layers, we use the following procedure:
Step 1. Obtain the times of first arrival, t1, t2, t3, c, at various distances
x1, x2, x3, c from the point of impact.
Step 2. Plot a graph of time t against distance x. The graph will look like the one
shown in Figure 2.42b.
Step 3. Determine the slopes of the lines ab, bc, cd, c:
Slope of ab 5
1
v1
Slope of bc 5
1
v2
Slope of cd 5
1
v3
Here, v1, v2, v3, c are the P-wave velocities in layers I, II, III, c,
respectively (Figure 2.42a).
Step 4. Determine the thickness of the top layer:
Z1 5
1 v2 2 v1
x
2 Å v2 1 v1 c
(2.73)
The value of xc can be obtained from the plot, as shown in Figure 2.42b.
Step 5. Determine the thickness of the second layer:
"v23 2 v21
v3v2
1
Z2 5 BTi2 2 2Z1
R
v3v1
2
"v23 2 v22
(2.74)
Here, Ti2 is the time intercept of the line cd in Figure 2.42b, extended backwards.
(For detailed derivatives of these equations and other related information, see Dobrin,
1960, and Das, 1992).
The velocities of P waves in various layers indicate the types of soil or rock that are
present below the ground surface. The range of the P-wave velocity that is generally
encountered in different types of soil and rock at shallow depths is given in Table 2.12.
In analyzing the results of a refraction survey, two limitations need to be kept in mind:
1. The basic equations for the survey—that is, Eqs. (2.73) and (2.74)—are based on the
assumption that the P-wave velocity v1 , v2 , v3 , c.
2. When a soil is saturated below the water table, the P-wave velocity may be deceptive.
P waves can travel with a velocity of about 1500 m>sec through water. For dry, loose
soils, the velocity may be well below 1500 m>sec. However, in a saturated condition,
the waves will travel through water that is present in the void spaces with a velocity
of about 1500 m>sec. If the presence of groundwater has not been detected, the
P-wave velocity may be erroneously interpreted to indicate a stronger material
(e.g., sandstone) than is actually present in situ. In general, geophysical interpretations should always be verified by the results obtained from borings.
2.26 Geophysical Exploration
121
Table 2.12 Range of P-Wave Velocity in Various Soils and Rocks
P-wave velocity
m , sec
Type of soil or rock
Soil
Sand, dry silt, and fine-grained topsoil
Alluvium
Compacted clays, clayey gravel, and
dense clayey sand
Loess
200–1000
500–2000
1000–2500
Slate and shale
Sandstone
Granite
Sound limestone
2500–5000
1500–5000
4000–6000
5000–10,000
250–750
Rock
Example 2.1
The results of a refraction survey at a site are given in the following table:
Distance of geophone from
the source of disturbance (m)
Time of first arrival
(sec 3 103)
2.5
5
7.5
10
15
20
25
30
35
40
50
11.2
23.3
33.5
42.4
50.9
57.2
64.4
68.6
71.1
72.1
75.5
Determine the P-wave velocities and the thickness of the material encountered.
Solution
Velocity
In Figure 2.43, the times of first arrival of the P waves are plotted against the distance of
the geophone from the source of disturbance. The plot has three straight-line segments.
The velocity of the top three layers can now be calculated as follows:
Slope of segment 0a 5
1
time
23 3 1023
5
5
v1
distance
5.25
122 Chapter 2: Natural Soil Deposits and Subsoil Exploration
80
Time of first arrival, t (103)—in seconds
c
b
Ti2 = 65 10–3 sec
3.5
14.75
60
13.5
a
11
23
xc = 10.5 m
40
20
5.25
0
0
10
20
30
Distance, x (m)
40
50
Figure 2.43 Plot of first arrival time of P wave versus distance
of geophone from source of disturbance
or
v1 5
5.25 3 103
5 228 m , sec (top layer)
23
Slope of segment ab 5
1
13.5 3 10 23
5
v2
11
or
v2 5
11 3 103
5 814.8 m , sec (middle layer)
13.5
Slope of segment bc 5
1
3.5 3 10 23
5
v3
14.75
or
v3 5 4214 m , sec (third layer)
Comparing the velocities obtained here with those given in Table 2.12 indicates that the
third layer is a rock layer.
Thickness of Layers
From Figure 2.43, xc 5 10.5 m, so
Z1 5
1 v2 2 v1
x
2Å v2 1 v1 c
Thus,
Z1 5
1 814.8 2 228
3 10.5 5 3.94 m
2Å 814.8 1 228
123
2.26 Geophysical Exploration
Again, from Eq. (2.74)
2Z1"v23 2 v21 (v3 ) (v2 )
1
Z2 5 BTi2 2
R
2
(v3v1 )
"v23 2 v22
The value of Ti2 (from Figure 2.43) is 65 3 10 23 sec. Hence,
2(3.94)"(4214) 2 2 (228) 2
(4214) (814.8)
1
Z2 5 B65 3 10 23 2
R
2
(4214) (228)
"(4214) 2 2 (814.8) 2
1
5 (0.065 2 0.0345)830.47 5 12.66 m
2
Thus, the rock layer lies at a depth of Z1 1 Z2 5 3.94 1 12.66 5 16.60 m from the
surface of the ground.
■
Cross-Hole Seismic Survey
The velocity of shear waves created as the result of an impact to a given layer of soil can
be effectively determined by the cross-hole seismic survey (Stokoe and Woods, 1972). The
principle of this technique is illustrated in Figure 2.44, which shows two holes drilled into
the ground a distance L apart. A vertical impulse is created at the bottom of one borehole
by means of an impulse rod. The shear waves thus generated are recorded by a vertically
sensitive transducer. The velocity of shear waves can be calculated as
vs 5
L
t
(2.75)
where t 5 travel time of the waves.
Impulse
Oscilloscope
Vertical velocity
transducer
Vertical
velocity
transducer
Shear wave
L
Figure 2.44 Cross-hole method
of seismic survey
124 Chapter 2: Natural Soil Deposits and Subsoil Exploration
The shear modulus Gs of the soil at the depth at which the test is taken can be determined from the relation
vs 5
Gs
Å (g>g)
or
Gs 5
v2s g
g
(2.76)
where
vs 5 velocity of shear waves
g 5 unit weight of soil
g 5 acceleration due to gravity
The shear modulus is useful in the design of foundations to support vibrating machinery
and the like.
Resistivity Survey
Another geophysical method for subsoil exploration is the electrical resistivity survey. The
electrical resistivity of any conducting material having a length L and an area of cross section A can be defined as
r5
RA
L
(2.77)
where R 5 electrical resistance.
The unit of resistivity is ohm-centimeter or ohm-meter. The resistivity of various
soils depends primarily on their moisture content and also on the concentration of dissolved ions in them. Saturated clays have a very low resistivity; dry soils and rocks have a
high resistivity. The range of resistivity generally encountered in various soils and rocks is
given in Table 2.13.
The most common procedure for measuring the electrical resistivity of a soil profile makes use of four electrodes driven into the ground and spaced equally along a
straight line. The procedure is generally referred to as the Wenner method (Figure 2.45a).
Table 2.13 Representative Values of Resistivity
Material
Resistivity
(ohm ? m)
Sand
Clays, saturated silt
Clayey sand
Gravel
Weathered rock
Sound rock
500–1500
0–100
200–500
1500–4000
1500–2500
.5000
2.26 Geophysical Exploration
125
I
V
d
d
d
Layer 1
Resistivity, ␳1
Z1
Layer 2
Resistivity, ␳2
(a)
␳
Slope ␳2
Slope ␳1
Z1
d
(b)
Figure 2.45 Electrical resistivity
survey: (a) Wenner method; (b) empirical method for determining resistivity
and thickness of each layer
The two outside electrodes are used to send an electrical current I (usually a dc current
with nonpolarizing potential electrodes) into the ground. The current is typically in the
range of 50 to 100 milliamperes. The voltage drop, V, is measured between the two inside
electrodes. If the soil profile is homogeneous, its electrical resistivity is
r5
2pdV
I
(2.78)
In most cases, the soil profile may consist of various layers with different resistivities, and Eq. (2.78) will yield the apparent resistivity. To obtain the actual resistivity of various layers and their thicknesses, one may use an empirical method that
involves conducting tests at various electrode spacings (i.e., d is changed). The sum of
the apparent resistivities, Sr, is plotted against the spacing d, as shown in Figure 2.45b.
The plot thus obtained has relatively straight segments, the slopes of which give the
resistivity of individual layers. The thicknesses of various layers can be estimated as
shown in Figure 2.45b.
The resistivity survey is particularly useful in locating gravel deposits within a finegrained soil.
126 Chapter 2: Natural Soil Deposits and Subsoil Exploration
2.27
Subsoil Exploration Report
At the end of all soil exploration programs, the soil and rock specimens collected in the
field are subject to visual observation and appropriate laboratory testing. (The basic soil
tests were described in Chapter 1.) After all the required information has been compiled,
a soil exploration report is prepared for use by the design office and for reference during
future construction work. Although the details and sequence of information in such reports
may vary to some degree, depending on the structure under consideration and the person
compiling the report, each report should include the following items:
1. A description of the scope of the investigation
2. A description of the proposed structure for which the subsoil exploration has been
conducted
3. A description of the location of the site, including any structures nearby, drainage
conditions, the nature of vegetation on the site and surrounding it, and any other features unique to the site
4. A description of the geological setting of the site
5. Details of the field exploration—that is, number of borings, depths of borings, types
of borings involved, and so on
6. A general description of the subsoil conditions, as determined from soil specimens
and from related laboratory tests, standard penetration resistance and cone penetration resistance, and so on
7. A description of the water-table conditions
8. Recommendations regarding the foundation, including the type of foundation recommended, the allowable bearing pressure, and any special construction procedure that
may be needed; alternative foundation design procedures should also be discussed in
this portion of the report
9. Conclusions and limitations of the investigations
The following graphical presentations should be attached to the report:
1. A site location map
2. A plan view of the location of the borings with respect to the proposed structures
and those nearby
3. Boring logs
4. Laboratory test results
5. Other special graphical presentations
The exploration reports should be well planned and documented, as they will help in
answering questions and solving foundation problems that may arise later during design
and construction.
Problems
2.1
2.2
For a Shelby tube, given: outside diameter 5 76.2 mm and inside diameter 73 mm
in. What is the area ratio of the tube?
A soil profile is shown in Figure P2.2 along with the standard penetration numbers
in the clay layer. Use Eqs. (2.8) and (2.9) to determine the variation of cu and OCR
with depth. What is the average value of cu and OCR?
Problems
1.5 m
Groundwater
table
1.5 m
N60
5
1.5 m
8
1.5 m
A
8
127
Dry sand
␥ 16.5 kN/m3
Sand
␥sat 19 kN/m3
Clay
␥sat 16.8 kN/m3
1.5 m
9
1.5 m
10
Sand
Figure P2.2
2.3
2.4
2.5
2.6
2.7
Following is the variation of the field standard penetration number (N60 ) in a sand
deposit:
Depth (m)
N60
1.5
3
4.5
6
7.9
9
6
8
9
8
13
14
The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand
from 0 to a depth of 6 m is 18 kN>m3, and the saturated unit weight of sand for depth
6 to 12 m is 20.2 kN>m3. Use the relationship of Skempton given in Eq. (2.12) to calculate the corrected penetration numbers.
For the soil profile described in Problem 2.3, estimate an average peak soil friction
angle. Use Eq. (2.28).
Repeat Problem 2.4 using Eq. (2.27).
Refer to Problem 2.3. Using Eq. (2.20), determine the average relative density
of sand.
The following table gives the variation of the field standard penetration number
(N60 ) in a sand deposit:
Depth (m)
N60
1.5
3.0
4.5
6.0
7.5
9.0
5
11
14
18
16
21
128 Chapter 2: Natural Soil Deposits and Subsoil Exploration
2.8
The groundwater table is located at a depth of 12 m. The dry unit weight of sand from
0 to a depth of 12 m is 17.6 kN>m3. Assume that the mean grain size (D50 ) of the
sand deposit to be about 0.8 mm. Estimate the variation of the relative density with
depth for sand. Use Eq. (2.21).
Following are the standard penetration numbers determined from a sandy soil in
the field:
Depth (m)
Unit weight of soil (kN , m 3)
N60
3.0
4.5
6.0
7.5
9.0
10.5
12.0
16.66
16.66
16.66
18.55
18.55
18.55
18.55
7
9
11
16
18
20
22
Using Eq. (2.27), determine the variation of the peak soil friction angle, fr.
Estimate an average value of fr for the design of a shallow foundation.
(Note: For depth greater than 6 m, the unit weight of soil is 18.55 kN>m3.)
2.9 Refer to Problem 2.8. Assume that the sand is clean and normally consolidated.
Estimate the average value of the modulus of elasticity between depths of 6 m
and 9 m.
2.10 Following are the details for a soil deposit in sand:
Depth (m)
3.0
4.5
6.0
2.11
2.12
2.13
2.14
Effective overburden
pressure (kN , m 2)
55
82
98
Field standard
penetration number, N60
9
11
12
Assume the uniformity coefficient (Cu ) of the sand to be 2.8 and an overconsolidation ratio (OCR) of 2. Estimate the average relative density of the sand between the
depth of 3 to 6 m. Use Eq. (2.19).
Refer to Figure P2.2. Vane shear tests were conducted in the clay layer. The vane
dimensions were 63.5 mm (D) 3 127 mm (H). For the test at A, the torque required
to cause failure was 0.051 N # m. For the clay, given: liquid limit 5 46 and plastic
limit 5 21. Estimate the undrained cohesion of the clay for use in the design by
using Bjerrum’s l relationship [Eq. (2.35a)].
Refer to Problem 2.11. Estimate the overconsolidation ratio of the clay. Use
Eqs. (2.37) and (2.38).
a. A vane shear test was conducted in a saturated clay. The height and diameter of the
vane were 101.6 mm and 50.8 mm, respectively. During the test, the maximum
torque applied was 23 lb-ft. Determine the undrained shear strength of the clay.
b. The clay soil described in part (a) has a liquid limit of 58 and a plastic limit of
29. What would be the corrected undrained shear strength of the clay for design
purposes? Use Bjerrum’s relationship for l [Eq. (2.35a)].
Refer to Problem 2.13. Determine the overconsolidation ratio for the clay. Use
Eqs. (2.37) and (2.40). Use s0r 5 64.2 kN>m2.
Problems
129
2.15 In a deposit of normally consolidated dry sand, a cone penetration test was conducted. Following are the results:
2.16
2.17
2.18
2.19
Depth
(m)
Point resistance of
cone, qc (MN , m2)
1.5
3.0
4.5
6.0
7.5
9.0
2.06
4.23
6.01
8.18
9.97
12.42
Assuming the dry unit weight of sand to be 16 kN>m3, estimate the average peak
friction angle, fr, of the sand. Use Eq. (2.48).
Refer to Problem 2.15. Using Eq. (2.46), determine the variation of the relative
density with depth.
In the soil profile shown in Figure P2.17, if the cone penetration resistance (qc) at A
(as determined by an electric friction-cone penetrometer) is 0.8 MN> m.2, estimate
a. The undrained cohesion, cu
b. The overconsolidation ratio, OCR
In a pressuremeter test in a soft saturated clay, the measuring cell volume
Vo 5 535 cm3, po 5 42.4 kN>m2, pf 5 326.5 kN>m2, vo 5 46 cm3, and vf 5
180 cm3. Assuming Poisson’s ratio (ms ) to be 0.5 and using Figure 2.32, calculate
the pressuremeter modulus (Ep ).
A dilatometer test was conducted in a clay deposit. The groundwater table was located
at a depth of 3 m below the surface. At a depth of 8 m below the surface, the contact
pressure (po ) was 280 kN>m2 and the expansion stress (p1 ) was 350 kN>m2.
Determine the following:
a. Coefficient of at-rest earth pressure, Ko
b. Overconsolidation ratio, OCR
c. Modulus of elasticity, Es
Assume sor at a depth of 8 m to be 95 kN>m2 and ms 5 0.35.
2m
Water table
Clay
␥ 18 kN/m3
Clay
␥sat 20 kN/m3
4m
A
Figure P2.17
130 Chapter 2: Natural Soil Deposits and Subsoil Exploration
2.20 A dilatometer test was conducted in a sand deposit at a depth of 6 m. The groundwater table was located at a depth of 2 m below the ground surface. Given, for the
sand: gd 5 14.5 kN>m3 and gsat 5 19.8 kN>m3. The contact stress during the test
was 260 kN>m2. Estimate the soil friction angle, fr.
2.21 The P-wave velocity in a soil is 1900 m>sec. Assuming Poisson’s ratio to be 0.32,
calculate the modulus of elasticity of the soil. Assume that the unit weight of soil
is 18 kN>m3.
2.22 The results of a refraction survey (Figure 2.42a) at a site are given in the following
table. Determine the thickness and the P-wave velocity of the materials encountered.
Distance from the source
of disturbance (m)
Time of first arrival of
P- waves (sec 3 103)
2.5
5.0
7.5
10.0
15.0
20.0
25.0
30.0
40.0
50.0
5.08
10.16
15.24
17.01
20.02
24.2
27.1
28.0
31.1
33.9
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3
3.1
Shallow Foundations: Ultimate
Bearing Capacity
Introduction
To perform satisfactorily, shallow foundations must have two main characteristics:
1. They have to be safe against overall shear failure in the soil that supports them.
2. They cannot undergo excessive displacement, or settlement. (The term excessive is
relative, because the degree of settlement allowed for a structure depends on several
considerations.)
The load per unit area of the foundation at which shear failure in soil occurs is called the
ultimate bearing capacity, which is the subject of this chapter.
3.2
General Concept
Consider a strip foundation with a width of B resting on the surface of a dense sand or
stiff cohesive soil, as shown in Figure 3.1a. Now, if a load is gradually applied to the
foundation, settlement will increase. The variation of the load per unit area on the foundation (q) with the foundation settlement is also shown in Figure 3.1a. At a certain
point—when the load per unit area equals qu—a sudden failure in the soil supporting the
foundation will take place, and the failure surface in the soil will extend to the ground
surface. This load per unit area, qu, is usually referred to as the ultimate bearing capacity of the foundation. When such sudden failure in soil takes place, it is called general
shear failure.
If the foundation under consideration rests on sand or clayey soil of medium compaction (Figure 3.1b), an increase in the load on the foundation will also be accompanied by an increase in settlement. However, in this case the failure surface in the soil will
gradually extend outward from the foundation, as shown by the solid lines in Figure
3.1b. When the load per unit area on the foundation equals qu(1), movement of the foundation will be accompanied by sudden jerks. A considerable movement of the foundation is then required for the failure surface in soil to extend to the ground surface (as
shown by the broken lines in the figure). The load per unit area at which this happens
is the ultimate bearing capacity, qu. Beyond that point, an increase in load will be
133
134 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Load/unit area, q
B
qu
Failure
surface
in soil
(a)
Settlement
Load/unit area, q
B
qu(1)
qu
Failure
surface
(b)
Settlement
Load/unit area, q
B
qu(1)
qu
Failure
surface
(c)
qu
Surface
footing
Settlement
Figure 3.1 Nature of bearing capacity failure in soil: (a) general shear failure: (b) local shear failure;
(c) punching shear failure (Redrawn after Vesic, 1973) (Vesic, A. S. (1973). “Analysis of Ultimate
Loads of Shallow Foundations,” Journal of Soil Mechanics and Foundations Division, American
Society of Civil Engineers, Vol. 99, No. SM1, pp. 45–73. With permission from ASCE.)
accompanied by a large increase in foundation settlement. The load per unit area of
the foundation, qu(1), is referred to as the first failure load (Vesic, 1963). Note that a
peak value of q is not realized in this type of failure, which is called the local shear
failure in soil.
If the foundation is supported by a fairly loose soil, the load–settlement plot will
be like the one in Figure 3.1c. In this case, the failure surface in soil will not extend
to the ground surface. Beyond the ultimate failure load, qu, the load–settlement plot
will be steep and practically linear. This type of failure in soil is called the punching
shear failure.
Vesic (1963) conducted several laboratory load-bearing tests on circular and rectangular plates supported by a sand at various relative densities of compaction, Dr. The variations of qu(1)> 12gB and qu> 12gB obtained from those tests, where B is the diameter of a
circular plate or width of a rectangular plate and g is a dry unit weight of sand, are shown
in Figure 3.2. It is important to note from this figure that, for Dr > about 70%, the general
shear type of failure in soil occurs.
On the basis of experimental results, Vesic (1973) proposed a relationship for
the mode of bearing capacity failure of foundations resting on sands. Figure 3.3
3.2 General Concept
0.2
0.3
0.4
Punching
shear
700
600
500
Relative density, Dr
0.5
0.6
0.7
0.8
135
0.9
General
shear
Local shear
400
300
qu
qu(1)
100
90
80
70
60
50
qu
1 B
2
and
1 B
2
200
1 B
2
40
Legend
Circular plate 203 mm (8 in.)
Circular plate 152 mm (6 in.)
Circular plate 102 mm (4 in.)
Circular plate 51 mm (2 in.)
Rectangular plate 51 305 mm
(2 12 in.)
30
qu(1)
1 B
2
20
Reduced by 0.6
Small signs indicate first failure load
10
1.32
1.35
1.40
1.45
1.50
Dry unit weight, d
Unit weight of water, w
1.55
1.60
Figure 3.2 Variation of qu(1)>0.5gB and qu>0.5gB for circular and rectangular plates on the
surface of a sand (Adapted from Vesic, 1963) (From Vesic, A. B. Bearing Capacity of Deep
Foundations in Sand. In Highway Research Record 39, Highway Research Board, National
Research Council, Washington, D.C., 1963, Figure 28, p. 137. Reproduced with permission of the
Transportation Research Board.)
shows this relationship, which involves the notation
Dr 5 relative density of sand
Df 5 depth of foundation measured from the ground surface
2BL
B* 5
B1L
where
B 5 width of foundation
L 5 length of foundation
(Note: L is always greater than B.)
(3.1)
136 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
0.2
0
Relative density, Dr
0.4
0.6
0.8
1.0
0
1
Punching shear
failure
Local shear
failure
General
shear
failure
Df /B*
2
3
4
Df
B
5
Figure 3.3 Modes of foundation failure in sand (After Vesic, 1973) (Vesic, A. S. (1973).
“Analysis of Ultimate Loads of Shallow Foundations,” Journal of Soil Mechanics and
Foundations Division, American Society of Civil Engineers, Vol. 99, No. SM1, pp. 45–73.
With permission from ASCE.)
For square foundations, B 5 L; for circular foundations, B 5 L 5 diameter, so
B* 5 B
(3.2)
Figure 3.4 shows the settlement S of the circular and rectangular plates on the surface of a
sand at ultimate load, as described in Figure 3.2. The figure indicates a general range of
S>B with the relative density of compaction of sand. So, in general, we can say that, for
foundations at a shallow depth (i.e., small Df> B*), the ultimate load may occur at a
foundation settlement of 4 to 10% of B. This condition arises together with general shear
failure in soil; however, in the case of local or punching shear failure, the ultimate load
may occur at settlements of 15 to 25% of the width of the foundation (B).
3.3
Terzaghi’s Bearing Capacity Theory
Terzaghi (1943) was the first to present a comprehensive theory for the evaluation of the
ultimate bearing capacity of rough shallow foundations. According to this theory, a foundation is shallow if its depth, Df (Figure 3.5), is less than or equal to its width. Later investigators, however, have suggested that foundations with Df equal to 3 to 4 times their
width may be defined as shallow foundations.
Terzaghi suggested that for a continuous, or strip, foundation (i.e., one whose widthto-length ratio approaches zero), the failure surface in soil at ultimate load may be assumed
to be similar to that shown in Figure 3.5. (Note that this is the case of general shear failure,
as defined in Figure 3.1a.) The effect of soil above the bottom of the foundation may also be
assumed to be replaced by an equivalent surcharge, q 5 gDf (where g is a unit weight of
soil). The failure zone under the foundation can be separated into three parts (see Figure 3.5):
3.3 Terzaghi’s Bearing Capacity Theory
0.3
0.2
25%
0.4
Relative density, Dr
0.5
0.6
Punching
shear
0.7
Local shear
0.8
General
shear
20%
S
B
Rectangular
plates
Circular plates
15%
10%
5%
Circular plate diameter
203 mm (8 in.)
152 mm (6 in.)
102 mm (4 in.)
51 mm (2 in.)
51 305 mm (2 12 in.)
Rectangular plate (width B)
0%
1.35
1.40
1.45
1.50
Dry unit weight, d
Unit weight of water, w
1.55
Figure 3.4 Range of settlement of circular and rectangular plates at ultimate
load (Df>B 5 0) in sand (Modified from Vesic, 1963) (From Vesic, A. B.
Bearing Capacity of Deep Foundations in Sand. In Highway Research Record
39, Highway Research Board, National Research Council, Washington, D.C.,
1963, Figure 29, p. 138. Reproduced with permission of the Transportation
Research Board.)
B
J
I
qu
Df
H
45 /2
A
45 /2
F
C
D
q Df
G
45 /2 45 /2
E Soil
Unit weight Cohesion
c
Friction angle Figure 3.5 Bearing capacity failure in soil under a rough rigid
continuous (strip) foundation
137
138 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
1. The triangular zone ACD immediately under the foundation
2. The radial shear zones ADF and CDE, with the curves DE and DF being arcs of
a logarithmic spiral
3. Two triangular Rankine passive zones AFH and CEG
The angles CAD and ACD are assumed to be equal to the soil friction angle fr.
Note that, with the replacement of the soil above the bottom of the foundation by an
equivalent surcharge q, the shear resistance of the soil along the failure surfaces GI and
HJ was neglected.
Using equilibrium analysis, Terzaghi expressed the ultimate bearing capacity in
the form
qu 5 crNc 1 qNq 1 12 gBNg
(continuous or strip foundation)
(3.3)
where
cr 5 cohesion of soil
g 5 unit weight of soil
q 5 gDf
Nc, Nq, Ng 5 bearing capacity factors that are nondimensional and are functions only of
the soil friction angle fr
The bearing capacity factors Nc, Nq, and Ng are defined by
Nc 5 cot fr
e2 (3p>42fr>2)tan fr
C
fr
p
2 cos ¢ 1 ≤
4
2
2
Nq 5
2 1 5 cot fr(Nq 2 1)
S
e2 (3p>42fr>2)tan fr
fr
2 cos ¢ 45 1 ≤
2
(3.4)
(3.5)
2
and
1 Kpg
Ng 5 ¢
2 1≤ tan fr
2 cos2 fr
(3.6)
where Kpg 5 passive pressure coefficient.
The variations of the bearing capacity factors defined by Eqs. (3.4), (3.5), and (3.6)
are given in Table 3.1.
To estimate the ultimate bearing capacity of square and circular foundations,
Eq. (3.1) may be respectively modified to
qu 5 1.3crNc 1 qNq 1 0.4gBNg
(square foundation)
(3.7)
3.3 Terzaghi’s Bearing Capacity Theory
139
Table 3.1 Terzaghi’s Bearing Capacity Factors—Eqs. (3.4), (3.5), and (3.6) a From
Kumbhojkar (1993)
a
f9
Nc
Nq
N ga
f9
Nc
Nq
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5.70
6.00
6.30
6.62
6.97
7.34
7.73
8.15
8.60
9.09
9.61
10.16
10.76
11.41
12.11
12.86
13.68
14.60
15.12
16.56
17.69
18.92
20.27
21.75
23.36
25.13
1.00
1.10
1.22
1.35
1.49
1.64
1.81
2.00
2.21
2.44
2.69
2.98
3.29
3.63
4.02
4.45
4.92
5.45
6.04
6.70
7.44
8.26
9.19
10.23
11.40
12.72
0.00
0.01
0.04
0.06
0.10
0.14
0.20
0.27
0.35
0.44
0.56
0.69
0.85
1.04
1.26
1.52
1.82
2.18
2.59
3.07
3.64
4.31
5.09
6.00
7.08
8.34
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
27.09
29.24
31.61
34.24
37.16
40.41
44.04
48.09
52.64
57.75
63.53
70.01
77.50
85.97
95.66
106.81
119.67
134.58
151.95
172.28
196.22
224.55
258.28
298.71
347.50
14.21
15.90
17.81
19.98
22.46
25.28
28.52
32.23
36.50
41.44
47.16
53.80
61.55
70.61
81.27
93.85
108.75
126.50
147.74
173.28
204.19
241.80
287.85
344.63
415.14
N ga
9.84
11.60
13.70
16.18
19.13
22.65
26.87
31.94
38.04
45.41
54.36
65.27
78.61
95.03
115.31
140.51
171.99
211.56
261.60
325.34
407.11
512.84
650.67
831.99
1072.80
From Kumbhojkar (1993)
and
qu 5 1.3crNc 1 qNq 1 0.3gBNg
(circular foundation)
(3.8)
In Eq. (3.7), B equals the dimension of each side of the foundation; in Eq. (3.8), B equals
the diameter of the foundation.
For foundations that exhibit the local shear failure mode in soils, Terzaghi suggested
the following modifications to Eqs. (3.3), (3.7), and (3.8):
qu 5 23crN cr 1 qN qr 1 12gBN gr
(strip foundation)
qu 5 0.867crN cr 1 qN qr 1 0.4gBN gr
(square foundation)
(3.10)
qu 5 0.867crN cr 1 qN qr 1 0.3gBN gr
(circular foundation)
(3.11)
(3.9)
140 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Table 3.2 Terzaghi’s Modified Bearing Capacity Factors Ncr, Nqr, and Ngr
f9
N 9c
N 9q
N 9g
f9
N 9c
N 9q
N 9g
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5.70
5.90
6.10
6.30
6.51
6.74
6.97
7.22
7.47
7.74
8.02
8.32
8.63
8.96
9.31
9.67
10.06
10.47
10.90
11.36
11.85
12.37
12.92
13.51
14.14
14.80
1.00
1.07
1.14
1.22
1.30
1.39
1.49
1.59
1.70
1.82
1.94
2.08
2.22
2.38
2.55
2.73
2.92
3.13
3.36
3.61
3.88
4.17
4.48
4.82
5.20
5.60
0.00
0.005
0.02
0.04
0.055
0.074
0.10
0.128
0.16
0.20
0.24
0.30
0.35
0.42
0.48
0.57
0.67
0.76
0.88
1.03
1.12
1.35
1.55
1.74
1.97
2.25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
15.53
16.30
17.13
18.03
18.99
20.03
21.16
22.39
23.72
25.18
26.77
28.51
30.43
32.53
34.87
37.45
40.33
43.54
47.13
51.17
55.73
60.91
66.80
73.55
81.31
6.05
6.54
7.07
7.66
8.31
9.03
9.82
10.69
11.67
12.75
13.97
15.32
16.85
18.56
20.50
22.70
25.21
28.06
31.34
35.11
39.48
44.45
50.46
57.41
65.60
2.59
2.88
3.29
3.76
4.39
4.83
5.51
6.32
7.22
8.35
9.41
10.90
12.75
14.71
17.22
19.75
22.50
26.25
30.40
36.00
41.70
49.30
59.25
71.45
85.75
Ncr, Nqr, and Ngr, the modified bearing capacity factors, can be calculated by using
the bearing capacity factor equations (for Nc, Nq, and Ng, respectively) by replacing fr
by fr 5 tan21 ( 23 tan fr). The variation of Ncr, Nqr, and Ngr with the soil friction angle fr
is given in Table 3.2.
Terzaghi’s bearing capacity equations have now been modified to take into account
the effects of the foundation shape (B>L), depth of embedment (Df ), and the load inclination. This is given in Section 3.6. Many design engineers, however, still use Terzaghi’s
equation, which provides fairly good results considering the uncertainty of the soil conditions at various sites.
3.4
Factor of Safety
Calculating the gross allowable load-bearing capacity of shallow foundations requires the
application of a factor of safety (FS) to the gross ultimate bearing capacity, or
qall 5
qu
FS
(3.12)
3.4 Factor of Safety
141
However, some practicing engineers prefer to use a factor of safety such that
Net stress increase on soil 5
net ultimate bearing capacity
FS
(3.13)
The net ultimate bearing capacity is defined as the ultimate pressure per unit area of the
foundation that can be supported by the soil in excess of the pressure caused by the
surrounding soil at the foundation level. If the difference between the unit weight of concrete used in the foundation and the unit weight of soil surrounding is assumed to be
negligible, then
qnet(u) 5 qu 2 q
(3.14)
where
qnet(u) 5 net ultimate bearing capacity
q 5 gDf
So
qall(net) 5
qu 2 q
FS
(3.15)
The factor of safety as defined by Eq. (3.15) should be at least 3 in all cases.
Example 3.1
A square foundation is 2 m 3 2 m in plan. The soil supporting the foundation has a
friction angle of fr 5 25° and cr 5 20 kN>m2. The unit weight of soil, g, is
16.5 kN>m3. Determine the allowable gross load on the foundation with a factor of
safety (FS) of 3. Assume that the depth of the foundation (Df ) is 1.5 m and that general
shear failure occurs in the soil.
Solution
From Eq. (3.7)
qu 5 1.3crNc 1 qNq 1 0.4gBNg
From Table 3.1, for fr 5 25°,
Nc 5 25.13
Nq 5 12.72
Ng 5 8.34
Thus,
qu 5 (1.3) (20) (25.13) 1 (1.5 3 16.5) (12.72) 1 (0.4) (16.5) (2) (8.34)
5 653.38 1 314.82 1 110.09 5 1078.29 kN>m2
142 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
So, the allowable load per unit area of the foundation is
qu
1078.29
5
< 359.5 kN>m2
qall 5
FS
3
Thus, the total allowable gross load is
Q 5 (359.5) B2 5 (359.5) (2 3 2) 5 1438 kN
3.5
■
Modification of Bearing Capacity Equations
for Water Table
Equations (3.3) and (3.7) through (3.11) give the ultimate bearing capacity, based on the
assumption that the water table is located well below the foundation. However, if the water
table is close to the foundation, some modifications of the bearing capacity equations will
be necessary. (See Figure 3.6.)
Case I. If the water table is located so that 0 # D1 # Df, the factor q in the bearing
capacity equations takes the form
(3.16)
q 5 effective surcharge 5 D1g 1 D2 (gsat 2 gw )
where
gsat 5 saturated unit weight of soil
gw 5 unit weight of water
Also, the value of g in the last term of the equations has to be replaced by gr 5 gsat 2 gw.
Case II. For a water table located so that 0 # d # B,
(3.17)
q 5 gDf
In this case, the factor g in the last term of the bearing capacity equations must be replaced
by the factor
g 5 gr 1
Groundwater
table
d
(g 2 gr)
B
(3.18)
D1
Case I
Df
D2
B
d
Groundwater table
Case II
sat saturated
unit weight
Figure 3.6 Modification of bearing
capacity equations for water table
3.6 The General Bearing Capacity Equation
143
The preceding modifications are based on the assumption that there is no seepage force in
the soil.
Case III. When the water table is located so that d $ B, the water will have no effect on
the ultimate bearing capacity.
3.6
The General Bearing Capacity Equation
The ultimate bearing capacity equations (3.3), (3.7), and (3.8) are for continuous, square,
and circular foundations only; they do not address the case of rectangular foundations
(0 , B>L , 1). Also, the equations do not take into account the shearing resistance
along the failure surface in soil above the bottom of the foundation (the portion of the failure surface marked as GI and HJ in Figure 3.5). In addition, the load on the foundation
may be inclined. To account for all these shortcomings, Meyerhof (1963) suggested the
following form of the general bearing capacity equation:
qu 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12 gBNgFgsFgdFgi
(3.19)
In this equation:
cr 5
q5
g5
B5
Fcs, Fqs, Fgs 5
Fcd, Fqd, Fgd 5
Fci, Fqi, Fgi 5
Nc, Nq, Ng 5
cohesion
effective stress at the level of the bottom of the foundation
unit weight of soil
width of foundation (5 diameter for a circular foundation)
shape factors
depth factors
load inclination factors
bearing capacity factors
The equations for determining the various factors given in Eq. (3.19) are described briefly
in the sections that follow. Note that the original equation for ultimate bearing capacity is
derived only for the plane-strain case (i.e., for continuous foundations). The shape, depth,
and load inclination factors are empirical factors based on experimental data.
Bearing Capacity Factors
The basic nature of the failure surface in soil suggested by Terzaghi now appears to have been
borne out by laboratory and field studies of bearing capacity (Vesic, 1973). However, the
angle a shown in Figure 3.5 is closer to 45 1 fr>2 than to fr. If this change is accepted, the
values of Nc, Nq, and Ng for a given soil friction angle will also change from those given in
Table 3.1. With a 5 45 1 fr>2, it can be shown that
Nq 5 tan2 ¢45 1
fr p tan fr
≤e
2
(3.20)
144 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
and
Nc 5 (Nq 2 1) cot fr
(3.21)
Equation (3.21) for Nc was originally derived by Prandtl (1921), and Eq. (3.20) for Nq was
presented by Reissner (1924). Caquot and Kerisel (1953) and Vesic (1973) gave the relation for Ng as
Ng 5 2 (Nq 1 1) tan fr
(3.22)
Table 3.3 shows the variation of the preceding bearing capacity factors with soil friction
angles.
Table 3.3 Bearing Capacity Factors
f9
Nc
Nq
Ng
f9
Nc
Nq
Ng
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
5.14
5.38
5.63
5.90
6.19
6.49
6.81
7.16
7.53
7.92
8.35
8.80
9.28
9.81
10.37
10.98
11.63
12.34
13.10
13.93
14.83
15.82
16.88
18.05
19.32
20.72
1.00
1.09
1.20
1.31
1.43
1.57
1.72
1.88
2.06
2.25
2.47
2.71
2.97
3.26
3.59
3.94
4.34
4.77
5.26
5.80
6.40
7.07
7.82
8.66
9.60
10.66
0.00
0.07
0.15
0.24
0.34
0.45
0.57
0.71
0.86
1.03
1.22
1.44
1.69
1.97
2.29
2.65
3.06
3.53
4.07
4.68
5.39
6.20
7.13
8.20
9.44
10.88
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
22.25
23.94
25.80
27.86
30.14
32.67
35.49
38.64
42.16
46.12
50.59
55.63
61.35
67.87
75.31
83.86
93.71
105.11
118.37
133.88
152.10
173.64
199.26
229.93
266.89
11.85
13.20
14.72
16.44
18.40
20.63
23.18
26.09
29.44
33.30
37.75
42.92
48.93
55.96
64.20
73.90
85.38
99.02
115.31
134.88
158.51
187.21
222.31
265.51
319.07
12.54
14.47
16.72
19.34
22.40
25.99
30.22
35.19
41.06
48.03
56.31
66.19
78.03
92.25
109.41
130.22
155.55
186.54
224.64
271.76
330.35
403.67
496.01
613.16
762.89
Shape, Depth, Inclination Factors
Commonly used shape, depth, and inclination factors are given in Table 3.4.
3.6 The General Bearing Capacity Equation
145
Table 3.4 Shape, Depth and Inclination Factors (DeBeer (1970); Hansen (1970); Meyerhof (1963);
Meyerhof and Hanna (1981))
Factor
Shape
Depth
Relationship
Reference
B Nq
Fcs 5 1 1 a b a b
L Nc
B
Fqs 5 1 1 a b tan fr
L
B
Fgs 5 120.4 a b
L
Df
1
B
For ␾ ⫽ 0:
Df
Fcd 5 1 1 0.4 a b
B
DeBeer (1970)
Hansen (1970)
Fqd ⫽ 1
F␥d ⫽ 1
For ␾⬘ ⬎ 0:
Fcd 5 Fqd 2
1 2 Fqd
Nc tan fr
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2 a
Df
B
b
F␥d ⫽ 1
Df
B
1
For ␾ ⫽ 0:
Df
Fcd 5 1 1 0.4 tan 21 a b
B
('')''*
radians
Fqd ⫽ 1
F␥d ⫽ 1
For ␾⬘ ⬎ 0:
Fcd 5 Fqd 2
1 2 Fqd
Nc tan f9
Df
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2 tan 21 a b
B
('')''*
radians
F␥d ⫽ 1
Inclination
Fci 5 Fqi 5 a1 2
Fgi 5 a1 2
b
f9
b° 2
b
90°
b
␤ ⫽ inclination of the load on the
foundation with respect to the vertical
Meyerhof (1963); Hanna and
Meyerhof (1981)
146 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Example 3.2
Solve Example Problem 3.1 using Eq. (3.19).
Solution
From Eq. (3.19),
qu 5 c9NcFcsFcdFci 1 qNqFqsFqdFqt 1
1
gBNgFgsFgdFgt
2
Since the load is vertical, Fci ⫽ Fqi ⫽ F␥i ⫽ 1. From Table 3.3 for ␾⬘ ⫽ 25°, Nc ⫽ 20.72,
Nq ⫽ 10.66, and N␥ ⫽ 10.88.
Using Table 3.4,
B Nq
2 10.66
Fcs 5 1 1 a b a b 5 1 1 a b a
b 5 1.514
L Nc
2 20.72
B
2
Fqs 5 1 1 a b tanf9 5 1 1 a b tan 25 5 1.466
L
2
B
2
Fgs 5 1 2 0.4a b 5 1 2 0.4a b 5 0.6
L
2
Fqd 5 1 1 2 tanfr (1 2 sinfr) 2 a
Df
B
b
5 1 1 (2) (tan 25) (1 2 sin 25) 2 a
Fcd 5 Fqd 2
1 2 Fqd
Nc tanf9
5 1.233 2 c
1.5
b 5 1.233
2
1 2 1.233
d 5 1.257
(20.72) (tan 25)
Fgd 5 1
Hence,
qu ⫽ (20)(20.72)(1.514)(1.257)(1)
⫹(1.5 ⫻ 16.5)(10.66)(1.466)(1.233)(1)
1
1 (16.5) (2) (10.88) (0.6) (1) (1)
2
⫽ 788.6 ⫹ 476.9 ⫹ 107.7 ⫽ 1373.2 kN/m2
qall 5
qu
1373.2
5
5 457.7 kN>m2
FS
3
Q ⫽ (457.7)(2 ⫻ 2) ⫽ 1830.8 kN
■
3.6 The General Bearing Capacity Equation
147
Example 3.3
A square foundation (B 3 B) has to be constructed as shown in Figure 3.7. Assume
that g 5 16.5 kN>m3, gsat 5 18.55 kN>m3, fr 5 34°, Df 5 1.22 m, and D1 5 0.61 m.
The gross allowable load, Qall, with FS 5 3 is 667.2 kN. Determine the size of the
footing. Use Eq. (3.19).
D1
Water
table
; ; c 0
sat
c 0
Df
BB
Figure 3.7 A square foundation
Solution
We have
qall 5
Qall
B2
5
667.2
kN>m2
B2
(a)
From Eq. (3.19) (with cr 5 0), for vertical loading, we obtain
qall 5
qu
1
1
5 ¢ qNqFqsFqd 1 grBNgFgsFgd ≤
FS
3
2
For fr 5 34°, from Table 3.3, Nq 5 29.44 and Ng 5 41.06. Hence,
Fqs 5 1 1
B
tan fr 5 1 1 tan 34 5 1.67
L
Fgs 5 1 2 0.4 ¢
B
≤ 5 1 2 0.4 5 0.6
L
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2
Df
B
5 1 1 2 tan 34 (1 2 sin 34) 2
4
1.05
511
B
B
Fgd 5 1
and
q 5 (0.61) (16.5) 1 0.61 (18.55 2 9.81) 5 15.4 kN>m2
148 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
So
qall 5
1
1.05
B(15.4) (29.44) (1.67) ¢1 1
≤
3
B
1
1 ¢ ≤ (18.55 2 9.81) (B) (41.06) (0.6) (1) R
2
5 252.38 1
(b)
265
1 35.89B
B
Combining Eqs. (a) and (b) results in
667.2
265
5 252.38 1
1 35.89B
2
B
B
By trial and error, we find that B < 1.3 m.
3.7
■
Case Studies on Ultimate Bearing Capacity
In this section, we will consider two field observations related to the ultimate bearing
capacity of foundations on soft clay. The failure loads on the foundations in the field will
be compared with those estimated from the theory presented in Section 3.6.
Foundation Failure of a Concrete Silo
An excellent case of bearing capacity failure of a 6-m diameter concrete silo was provided by Bozozuk (1972). The concrete tower silo was 21 m high and was constructed
over soft clay on a ring foundation. Figure 3.8 shows the variation of the undrained shear
strength (cu) obtained from field vane shear tests at the site. The groundwater table was
located at about 0.6 m below the ground surface.
On September 30, 1970, just after it was filled to capacity for the first time with
corn silage, the concrete tower silo suddenly overturned due to bearing capacity
failure. Figure 3.9 shows the approximate profile of the failure surface in soil. The
failure surface extended to about 7 m below the ground surface. Bozozuk (1972)
provided the following average parameters for the soil in the failure zone and the
foundation:
•
•
•
•
Load per unit area on the foundation when failure occurred < 160 kN>m2
Average plasticity index of clay (PI) < 36
Average undrained shear strength (cu) from 0.6 to 7 m depth obtained from field vane
shear tests < 27.1 kN>m2
From Figure 3.9, B < 7.2 m and Df < 1.52 m.
cu (kN/m2)
40
60
20
0
80
100
1
Depth (m)
2
3
4
5
Figure 3.8 Variation of cu with depth
obtained from field vane shear test
6
50
Original position
of foundation
1.46 m
0
22 2 1m
m
30
0.9
4
6
Paved apron
Original
ground surface
7.2
Depth below paved apron (m)
1
2
Collapsed silo
50
Upheaval
22
45
m
1.
60
8
10
12
Figure 3.9 Approximate profile of silo failure (Adapted from Bozozuk, 1972)
149
150 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
We can now calculate the factor of safety against bearing capacity failure. From Eq. (3.19)
qu 5 crNcFcsFcdFci 1 qNcFqsFqdFqi 1 12 gB NgFgsFgdFgi
For f 5 0 condition and vertical loading, cr 5 cu, Nc 5 5.14, Nq 5 1, Ng 5 0, and
Fci 5 Fqi 5 Fgi 5 0. Also, from Table 3.4,
Fcs 5 1 1 a
7.2
1
ba
b 5 1.195
7.2 5.14
Fqs 5 1
Fcd 5 1 1 (0.4) a
1.52
b 5 1.08
7.2
Fqd 5 1
Thus,
qu 5 (cu ) (5.14) (1.195) (1.08) (1) 1 (g) (1.52)
Assuming g < 18 kN>m3,
qu 5 6.63cu 1 27.36
(3.23)
According to Eqs. (2.34) and (2.35a),
cu(corrected) 5 l cu(VST)
l 5 1.7 2 0.54 log 3PI(%)4
For this case, PI < 36 and cu(VST) 5 27.1 kN>m2. So
cu(corrected) 5 51.7 2 0.54 log 3PI(%)46cu(VST)
5 (1.7 2 0.54 log 36) (27.1) < 23.3 kN>m2
Substituting this value of cu in Eq. (3.23)
qu 5 (6.63) (23.3) 1 27.36 5 181.8 kN>m2
The factor of safety against bearing capacity failure
FS 5
qu
181.8
5
5 1.14
applied load per unit area
160
This factor of safety is too low and approximately equals one, for which the failure occurred.
Load Tests on Small Foundations in Soft Bangkok Clay
Brand et al. (1972) reported load test results for five small square foundations in soft
Bangkok clay in Rangsit, Thailand. The foundations were 0.6 m ⫻ 0.6 m, 0.675 m ⫻
0.675 m, 0.75 m ⫻ 0.75 m, 0.9 m ⫻ 0.9 m, and 1.05 m ⫻ 1.05 m. The depth of of the
foundations (Df) was 1.5 m in all cases.
Figure 3.10 shows the vane shear test results for clay. Based on the variation of
cu(VST) with depth, it can be approximated that cu(VST) is about 35 kN/m2 for depths between
zero to 1.5 m measured from the ground surface, and cu(VST) is approximately equal to
24 kN/m2 for depths varying from 1.5 to 8 m. Other properties of the clay are
•
•
•
Liquid limit ⫽ 80
Plastic limit ⫽ 40
Sensitivity ⬇ 5
3.7 Case Studies on Ultimate Bearing Capacity
151
cu (VST) (kN/m2)
0
10
20
30
40
1
2
Depth (m)
3
4
5
6
7
8
Figure 3.10 Variation of cu(VST) with depth for
soft Bangkok clay
Figure 3.11 shows the load-settlement plots obtained from the bearing-capacity tests on
all five foundations. The ultimate loads, Qu, obtained from each test are shown in Figure 3.11
and given in Table 3.5. The ultimate load is defined as the point where the load-settlement plot
becomes practically linear.
From Eq. (3.19),
qu 5 c9NcFcsFcdFci 1 qNqFqsFqdFqi 1
1
gBNgFgsFgdFgi
2
For undrained condition and vertical loading (that is, ␾ ⫽ 0) from Tables 3.3 and 3.4,
•
•
•
•
•
•
Fci ⫽ Fqi ⫽ F␥i ⫽ 1
c⬘ ⫽ cu, Nc ⫽ 5.14, Nq ⫽ 1, and N␥ ⫽ 0
B Nq
1
Fcs 5 1 1 a b a b 5 1 1 (1) a
b 5 1.195
L Nc
5.14
Fqs ⫽ 1
Fqd ⫽ 1
Df
1.5
Fcd 5 1 1 0.4 tan 21 a b 5 1 1 0.4 tan 21 a b
B
B
(3.24)
152 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Load (kN)
0
40
0
200
160
120
80
Qu (ultimate load)
Settlement (mm)
10
20
B = 0.675 m
30
B = 1.05 m
B = 0.6 m
B = 0.75 m
B = 0.9 m
40
Figure 3.11 Load-settlement plots obtained from bearing capacity tests
(Note: Df /B ⬎ 1 in all cases)
Thus,
qu ⫽ (5.14)(cu)(1.195)Fcd ⫹ q
(3.25)
The values of cu(VST) need to be corrected for use in Eq. (3.25). From Eq. (2.34),
cu ⫽ ␭cu(VST)
From Eq. (2.35b),
␭ ⫽ 1.18e⫺0.08(PI) ⫹ 0.57 ⫽ 1.18e⫺0.08(80 ⫺ 40) ⫹ 0.57 ⫽ 0.62
From Eq. (2.35c),
␭ ⫽ 7.01e⫺0.08(LL) ⫹ 0.57 ⫽ 7.01e⫺0.08(80) ⫹ 0.57 ⫽ 0.58
Table 3.5 Comparison of Ultimate Bearing Capacity—Theory versus Field Test Results
B
(m)
(1)
Df
(m)
(2)
0.600
0.675
0.750
0.900
1.050
1.5
1.5
1.5
1.5
1.5
Fcd
(3)
qu(theory)‡‡
(kN/m2)
(4)
1.476
1.459
1.443
1.412
1.384
158.3
156.8
155.4
152.6
150.16
‡
Qu(field) (kN)
(5)
qu(field)‡‡‡
(kN/m2)
(6)
qu(field) 2 qu(theory)
(%)
qu(field)
60
71
90
124
140
166.6
155.8
160.6
153.0
127.0
4.98
⫺0.64
2.87
0.27
⫺18.24
Eq. (3.24); ‡‡Eq. (3.26); ‡‡‡Qu(field)/B2 ⫽ qu(field)
‡
(7)
3.8 Effect of Soil Compressibility
153
So the average value of ␭ ⬇ 0.6. Hence,
cu ⫽ ␭cu(VST) ⫽ (0.6)(24) ⫽ 14.4 kN/m2
Let us assume ␥ ⫽ 18.5 kN/m2. So
q ⫽ ␥Df ⫽ (18.5)(1.5) ⫽ 27.75 kN/m2
Substituting cu ⫽ 14.4 kN/m2 and q ⫽ 27.75 kN/m2 into Eq. (3.25), we obtain
qu(kN/m2) ⫽ 88.4Fcd ⫹ 27.75
(3.26)
The values of qu calculated using Eq. (3.26) are given in column 4 of Table 3.5. Also,
the qu determined from the field tests are given in column 6. The theoretical and field values of qu compare very well. The important lessons learned from this study are
1. The ultimate bearing capacity is a function of cu. If Eq. (2.35a) would have been
used to correct the undrained shear strength, the theoretical values of qu would have
varied between 200 kN/m2 and 210 kN/m2. These values are about 25% to 55%
more than those obtained from the field and are on the unsafe side.
2. It is important to recognize that empirical correlations like those given in Eqs.
(2.35a), (2.35b) and (2.35c) are sometimes site specific. Thus, proper engineering
judgment and any record of past studies would be helpful in the evaluation of bearing capacity.
3.8
Effect of Soil Compressibility
In Section 3.3, Eqs. (3.3), (3.7), and (3.8), which apply to the case of general shear failure,
were modified to Eqs. (3.9), (3.10), and (3.11) to take into account the change of failure
mode in soil (i.e., local shear failure). The change of failure mode is due to soil compressibility, to account for which Vesic (1973) proposed the following modification of Eq. (3.19):
qu 5 crNcFcsFcdFcc 1 qNqFqsFqdFqc 1 12 gBNgFgsFgdFgc
(3.27)
In this equation, Fcc, Fqc, and Fgc are soil compressibility factors.
The soil compressibility factors were derived by Vesic (1973) by analogy to the
expansion of cavities. According to that theory, in order to calculate Fcc, Fqc, and Fgc, the
following steps should be taken:
Step 1. Calculate the rigidity index, Ir, of the soil at a depth approximately B>2
below the bottom of the foundation, or
Ir 5
Gs
cr 1 qr tan fr
where
Gs 5 shear modulus of the soil
q 5 effective overburden pressure at a depth of Df 1 B>2
(3.28)
154 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Step 2.
The critical rigidity index, Ir(cr), can be expressed as
fr
1
B
Ir(cr) 5 bexp B ¢ 3.30 2 0.45 ≤ cot ¢ 45 2 ≤ R r
2
L
2
Step 3.
(3.29)
The variations of Ir(cr) with B>L are given in Table 3.6.
If Ir $ Ir(cr), then
Fcc 5 Fqc 5 Fgc 5 1
However, if Ir , Ir(cr), then
Fgc 5 Fqc 5 exp b ¢ 24.4 1 0.6
(3.07 sin fr) (log 2Ir )
B
≤ tan fr 1 B
Rr
L
1 1 sin fr
(3.30)
Figure 3.12 shows the variation of Fgc 5 Fqc [see Eq. (3.30)] with fr and Ir. For f 5 0,
Fcc 5 0.32 1 0.12
B
1 0.60 log Ir
L
(3.31)
For fr . 0,
Fcc 5 Fqc 2
1 2 Fqc
(3.32)
Nq tan fr
Table 3.6 Variation of Ir(cr) with ␾⬘ and B/L
Ir(cr)
␾⬘
(deg)
B/L ⴝ 0
B/L ⴝ 0.2
B/L ⴝ 0.4
B/L ⴝ 0.6
B/L ⴝ 0.8
B/L ⴝ 1.0
0
5
10
15
20
25
30
35
40
45
13.56
18.30
25.53
36.85
55.66
88.93
151.78
283.20
593.09
1440.94
12.39
16.59
22.93
32.77
48.95
77.21
129.88
238.24
488.97
1159.56
11.32
15.04
20.60
29.14
43.04
67.04
111.13
200.41
403.13
933.19
10.35
13.63
18.50
25.92
37.85
58.20
95.09
168.59
332.35
750.90
9.46
12.36
16.62
23.05
33.29
50.53
81.36
141.82
274.01
604.26
8.64
11.20
14.93
20.49
29.27
43.88
69.62
119.31
225.90
486.26
3.8 Effect of Soil Compressibility
1.0
1.0
500
100 250
0.8
250
0.8
50
50
0.6
10
0.4
100
F c Fqc
F c Fqc
155
25
5
2.5
0.2
0.6
25
10
0.4
5
2.5
0.2
Ir 1
0
500
Ir 1
0
0
10
20
30
50
40
Soil friction angle, (deg)
L
(a)
1
B
0
10
20
30
50
40
Soil friction angle, (deg)
L
(b)
5
B
Figure 3.12 Variation of Fgc 5 Fqc with Ir and fr
Example 3.4
For a shallow foundation, B 5 0.6 m, L 5 1.2 m, and Df 5 0.6 m. The known soil
characteristics are as follows:
Soil:
fr 5 25°
cr 5 48 kN>m2
g 5 18 kN>m3
Modulus of elasticity, Es 5 620 kN>m2
Poisson’s ratio, ms 5 0.3
Calculate the ultimate bearing capacity.
Solution
From Eq. (3.28),
Ir 5
Gs
cr 1 qr tan fr
However,
Gs 5
Es
2 (1 1 ms )
So
Ir 5
Now,
qr 5 g ¢ Df 1
Es
2 (1 1 ms ) 3cr 1 qr tan fr4
B
0.6
≤ 5 18 ¢0.6 1
≤ 5 16.2 kN>m2
2
2
156 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Thus,
Ir 5
From Eq. (3.29),
620
5 4.29
2 (1 1 0.3) 348 1 16.2 tan 254
fr
1
B
Ir(cr) 5 bexp B ¢3.3 2 0.45 ≤ cot ¢45 2 ≤ R r
2
L
2
1
25
0.6
5 bexp B ¢ 3.3 2 0.45
≤ cot ¢45 2 ≤ R r 5 62.41
2
1.2
2
Since Ir(cr) . Ir, we use Eqs. (3.30) and (3.32) to obtain
Fgc 5 Fqc 5 exp b ¢ 24.4 1 0.6
5 exp b ¢ 24.4 1 0.6
1 c
(3.07 sin fr)log(2Ir )
B
≤ tan fr 1 B
Rr
L
1 1 sin fr
0.6
≤ tan 25
1.2
(3.07 sin 25)log(2 3 4.29)
R r 5 0.347
1 1 sin 25
and
Fcc 5 Fqc 2
1 2 Fqc
Nc tan fr
For fr 5 25°, Nc 5 20.72 (see Table 3.3); therefore,
1 2 0.347
Fcc 5 0.347 2
5 0.279
20.72 tan 25
Now, from Eq. (3.27),
qu 5 crNcFcsFcdFcc 1 qNqFqsFqdFqc 1 12gBNgFgsFgdFgc
From Table 3.3, for fr 5 25°, Nc 5 20.72, Nq 5 10.66, and Ng 5 10.88. Consequently,
Nq B
10.66 0.6
Fcs 5 1 1 ¢ ≤ ¢ ≤ 5 1 1 ¢
≤¢
≤ 5 1.257
Nc
L
20.72 1.2
Fqs 5 1 1
B
0.6
tan fr 5 1 1
tan 25 5 1.233
L
1.2
Fgs 5 1 2 0.4 ¢
B
0.6
≤ 5 1 2 0.4
5 0.8
L
1.2
Fqd 5 1 1 2 tan fr (1 2 sin fr) 2 ¢
Df
5 1 1 2 tan 25 (1 2 sin 25) 2 ¢
Fcd 5 Fqd 2
1 2 Fqd
Nc tan fr
5 1.311 2
B
≤
0.6
≤ 5 1.311
0.6
1 2 1.311
5 1.343
20.72 tan 25
3.9 Eccentrically Loaded Foundations
157
and
Fgd 5 1
Thus,
qu 5 (48) (20.72) (1.257) (1.343) (0.279) 1 (0.6 3 18) (10.66) (1.233) (1.311)
(0.347)1( 12 ) (18) (0.6) (10.88) (0.8) (1) (0.347) 5 549.32 kN , m2
3.9
■
Eccentrically Loaded Foundations
In several instances, as with the base of a retaining wall, foundations are subjected to
moments in addition to the vertical load, as shown in Figure 3.13a. In such cases, the distribution of pressure by the foundation on the soil is not uniform. The nominal distribution
of pressure is
qmax 5
Q
6M
1 2
BL
BL
(3.33)
qmin 5
Q
6M
2 2
BL
BL
(3.34)
and
Q
Q
e
M
B
B
BL
For e < B/6
qmin
qmax
L
For e > B/6
qmax
(a)
Figure 3.13 Eccentrically loaded foundations
2e
B
(b)
158 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
where
Q 5 total vertical load
M 5 moment on the foundation
Figure 3.13b shows a force system equivalent to that shown in Figure 3.13a. The
distance
e5
M
Q
(3.35)
is the eccentricity. Substituting Eq. (3.35) into Eqs. (3.33) and (3.34) gives
qmax 5
Q
6e
¢1 1 ≤
BL
B
(3.36)
qmin 5
Q
6e
¢1 2 ≤
BL
B
(3.37)
and
Note that, in these equations, when the eccentricity e becomes B>6, qmin is zero. For
e . B>6, qmin will be negative, which means that tension will develop. Because soil cannot
take any tension, there will then be a separation between the foundation and the soil underlying it. The nature of the pressure distribution on the soil will be as shown in Figure 3.13a.
The value of qmax is then
qmax 5
4Q
3L(B 2 2e)
(3.38)
The exact distribution of pressure is difficult to estimate.
Figure 3.14 shows the nature of failure surface in soil for a surface strip foundation
subjected to an eccentric load. The factor of safety for such type of loading against bearing capacity failure can be evaluated as
FS 5
Qult
Q
(3.39)
where Qult 5 ultimate load-carrying capacity.
The following sections describe several theories for determining Qult.
Qult
e
B
Figure 3.14 Nature of
failure surface in soil
supporting a strip foundation subjected to
eccentric loading
(Note: Df 5 0; Qult is ultimate load per unit length
of foundation)
3.10 Ultimate Bearing Capacity under Eccentric Loading—One-Way Eccentricity
3.10
159
Ultimate Bearing Capacity under Eccentric
Loading—One-Way Eccentricity
Effective Area Method (Meyerhoff, 1953)
In 1953, Meyerhof proposed a theory that is generally referred to as the effective area method.
The following is a step-by-step procedure for determining the ultimate load that the
soil can support and the factor of safety against bearing capacity failure:
Step 1.
Determine the effective dimensions of the foundation (Figure 3.13b):
Br 5 effective width 5 B 2 2e
Lr 5 effective length 5 L
Note that if the eccentricity were in the direction of the length of the foundation, the value of Lr would be equal to L 2 2e. The value of Br would
equal B. The smaller of the two dimensions (i.e., Lr and Br) is the effective
width of the foundation.
Step 2. Use Eq. (3.19) for the ultimate bearing capacity:
qur 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12 gBrNgFgsFgdFgi
(3.40)
To evaluate Fcs, Fqs, and Fgs, use the relationships given in Table 3.4 with
effective length and effective width dimensions instead of L and B, respectively. To determine Fcd, Fqd, and Fgd, use the relationships given in Table
3.4. However, do not replace B with Br.
Step 3. The total ultimate load that the foundation can sustain is
Ar &
$'%'
Qult 5
(3.41)
q ru (Br) (Lr)
where Ar 5 effective area.
Step 4. The factor of safety against bearing capacity failure is
Qult
FS 5
Q
Prakash and Saran Theory
Prakash and Saran (1971) analyzed the problem of ultimate bearing capacity of eccentrically and vertically loaded continuous (strip) foundations by using the one-sided failure
surface in soil, as shown in Figure 3.14. According to this theory, the ultimate load per unit
length of a continuous foundation can be estimated as
1
Q ult 5 B ccrNc(e) 1 qNq(e) 1 gBNg(e) d
2
(3.42)
where Nc(e), Nq(e), N␥(e) ⫽ bearing capacity factors under eccentric loading.
The variations of Nc(e), Nq(e), and N␥(e) with soil friction angle ␾⬘ are given in
Figures 3.15, 3.16, and 3.17. For rectangular foundations, the ultimate load can be given as
1
Q ult 5 BL ccrNc(e)Fcs(e) 1 qNq(e)Fqs(e) 1 gBNg(e)Fgs(e) d
2
where Fcs(e), Fqs(e), and F␥s(e) ⫽ shape factors.
(3.43)
160 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
60
40
e/B = 0
Nc (e)
0.1
0.2
20
0.3
0.4
f9 5 408
0
0
10
20
Friction angle, (deg)
30
40
e/B
Nc(e)
0
0.1
0.2
0.3
0.4
94.83
66.60
54.45
36.3
18.15
Figure 3.15 Variation of Nc(e) with fr
Prakash and Saran (1971) also recommended the following for the shape factors:
Fcs(e) 5 1.2 2 0.025
L
(with a minimum of 1.0)
B
Fqs(e) 5 1
(3.44)
(3.45)
and
Fgs(e) 5 1.0 1 a
2e
B
3 e
B 2
2 0.68b 1 c0.43 2 a b a b d a b
B
L
2 B
L
(3.46)
Reduction Factor Method (For Granular Soil)
Purkayastha and Char (1977) carried out stability analysis of eccentrically loaded continuous foundations supported by a layer of sand using the method of slices. Based on that
analysis, they proposed
3.10 Ultimate Bearing Capacity under Eccentric Loading—One-Way Eccentricity
161
60
40
e/B = 0
Nq (e)
0.1
20
0.2
0.3
f9 5 408
0.4
0
0
10
20
Friction angle, (deg)
30
40
e/B
Nq(e)
0
0.1
0.2
0.3
0.4
81.27
56.09
45.18
30.18
15.06
Figure 3.16 Variation of Nq(e) with fr
Rk 5 1 2
qu(eccentric)
qu(centric)
(3.47)
where
Rk ⫽ reduction factor
qu(eccentric) ⫽ ultimate bearing capacity of eccentrically loaded continuous
foundations
qu(centric) ⫽ ultimate bearing capacity of centrally loaded continuous foundations
The magnitude of Rk can be expressed as
e k
Rk 5 aa b
B
where a and k are functions of the embedment ratio Df /B (Table 3.7).
(3.48)
162 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
60
40
e/B = 0
N(e)
0.1
20
0.2
f9 5 408
0.3
0.4
0
0
10
20
Friction angle, (deg)
30
40
e,B
Ng(e)
0
0.1
0.2
0.3
0.4
115.80
71.80
41.60
18.50
4.62
Figure 3.17 Variation of Ng(e) with fr
Table 3.7 Variations of a and k [Eq. (3.48)]
Df/B
a
k
0.00
0.25
0.50
1.00
1.862
1.811
1.754
1.820
0.73
0.785
0.80
0.888
Hence, combining Eqs. (3.47) and (3.48)
e k
qu(eccentric) 5 qu(centric) (1 2 Rk ) 5 qu(centric) c1 2 aa b d
B
(3.49)
1
qu(centric) 5 qNqFqd 1 gBNgFgd
2
(3.50)
where
3.10 Ultimate Bearing Capacity under Eccentric Loading—One-Way Eccentricity
163
The relationships for Fqd, and F␥d are given in Table 3.4.
The ultimate load per unit length of the foundation can then be given as
Qu ⫽ Bqu(eccentric)
(3.51)
Example 3.5
A continuous foundation is shown in Figure 3.18. If the load eccentricity is 0.2 m,
determine the ultimate load, Qult, per unit length of the foundation. Use Meyerhof’s
effective area method.
Solution
For c⬘ ⫽ 0, Eq. (3.40) gives
qu9 5 qNqFqsFqdFqi 1
1
grBrNgFgsFgdFgi
2
where q ⫽ (16.5) (1.5) ⫽ 24.75 kN/m2.
Sand
40
c 0
16.5 kN/m3
1.5 m
2m
Figure 3.18 A continuous foundation with load
eccentricity
For ␾⬘ ⫽ 40°, from Table 3.3, Nq ⫽ 64.2 and N␥ ⫽ 109.41. Also,
B⬘ ⫽ 2 ⫺ (2)(0.2) ⫽ 1.6 m
Because the foundation in question is a continuous foundation, B⬘/L⬘ is zero. Hence,
Fqs ⫽ 1, F␥s ⫽ 1. From Table 3.4,
Fqi ⫽ F␥ i ⫽ 1
Fqd 5 1 1 2 tan fr(1 2 sin fr) 2
Df
B
5 1 1 0.214a
1.5
b 5 1.16
2
F␥d ⫽ 1
and
qur 5 (24.75) (64.2) (1) (1.16) (1)
1
1 a b (16.5) (1.6) (109.41) (1) (1) (1) 5 3287.39 kN>m2
2
Consequently,
Qult ⫽ (B⬘)(1)(qu⬘) ⫽ (1.6)(1)(3287.39) ⬇ 5260 kN
■
164 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Example 3.6
Solve Example 3.5 using Eq. (3.42).
Solution
Since c⬘ ⫽ 0
1
Qult 5 B cqNq(e) 1 gBNg(e) d
2
e
0.2
5
5 0.1
B
2
For ␾⬘ ⫽ 40° and e/B ⫽ 0.1, Figures 3.16 and 3.17 give Nq(e) ⫽ 56.09 and N␥ (e) ⬇ 71.8.
Hence,
Qult ⫽ 2[(24.75)(56.09) ⫹ (12)(16.5)(2)(71.8)] ⫽ 5146 kN
■
Example 3.7
Solve Example 3.5 using Eq. (3.49).
Solution
With c⬘ ⫽ 0,
1
qu(centric) 5 qNqFqd 1 gBNgFgd
2
For ␾⬘ ⫽ 40°, Nq ⫽ 64.2 and N␥ ⫽ 109.41 (see Table 3.3). Hence,
Fqd ⫽ 1.16 and F␥d ⫽ 1 (see Example 3.5)
1
qu(centric) 5 (24.75) (64.2) (1.16) 1 (16.5) (2) (109.41) (1)
2
5 1843.18 1 1805.27 5 3648.45 kN>m2
From Eq. (3.48),
e k
Rk 5 aa b
B
For Df /B ⫽ 1.5/2 ⫽ 0.75, Table 3.7 gives a ⬇ 1.75 and k ⬇ 0.85. Hence,
Rk 5 1.79a
0.2 0.85
b
5 0.253
2
Qu ⫽ Bqu(eccentric) ⫽ Bqu(centric)(1 ⫺ Rk) ⫽ (2)(3648.45)(1 ⫺ 0.253) ⬇ 5451 kN
■
3.11 Bearing Capacity—Two-way Eccentricity
3.11
165
Bearing Capacity—Two-way Eccentricity
Consider a situation in which a foundation is subjected to a vertical ultimate load Qult and
a moment M, as shown in Figures 3.19a and b. For this case, the components of the
moment M about the x- and y-axes can be determined as Mx and My, respectively. (See
Figure 3.19.) This condition is equivalent to a load Qult placed eccentrically on the foundation with x 5 eB and y 5 eL (Figure 3.19d). Note that
My
eB 5
(3.52)
Qult
and
eL 5
Mx
Qult
(3.53)
If Qult is needed, it can be obtained from Eq. (3.41); that is,
Qult 5 qur Ar
where, from Eq. (3.40),
qur 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12 gBrNgFgsFgdFgi
and
Ar 5 effective area 5 BrLr
As before, to evaluate Fcs, Fqs, and Fgs (Table 3.4), we use the effective length Lr and
effective width Br instead of L and B, respectively. To calculate Fcd, Fqd, and Fgd, we do
Qult
M
(a)
BL
B
y
eB
Mx
L
Qult
M
Qult
x
Qult
eL
My
B
(b)
(c)
(d)
Figure 3.19 Analysis of foundation with two-way eccentricity
166 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
not replace B with Br. In determining the effective area Ar, effective width Br, and effective length Lr, five possible cases may arise (Highter and Anders, 1985).
Case I. eL>L $ 16 and eB>B $ 16. The effective area for this condition is shown in
Figure 3.20, or
Ar 5 12B1L1
(3.54)
where
B1 5 B ¢1.5 2
3eB
≤
B
(3.55)
L1 5 L ¢1.5 2
3eL
≤
L
(3.56)
and
The effective length Lr is the larger of the two dimensions B1 and L1. So the effective width is
Br 5
Ar
Lr
(3.57)
Case II. eL>L , 0.5 and 0 , eB>B , 16. The effective area for this case, shown in
Figure 3.21a, is
Ar 5 12 (L1 1 L2 )B
(3.58)
The magnitudes of L1 and L2 can be determined from Figure 3.21b. The effective width is
L1 or L2
Ar
(whichever is larger)
(3.59)
Lr 5 L1 or L2
(whichever is larger)
(3.60)
Br 5
The effective length is
Case III. eL>L , 16 and 0 , eB>B , 0.5. The effective area, shown in Figure 3.22a, is
Ar 5 12 (B1 1 B2 )L
Effective
area
B1
eB
eL
Qult
L1
L
B
Figure 3.20 Effective area for the case of eL>L > 16
and eB>B > 16
(3.61)
3.11 Bearing Capacity–Two-way Eccentricity
167
Effective
area
B
eB
L2
eL
Qult
L1
L
(a)
0.5
eB /B 0.4
0.167
0.1
0.08
0.06
eL /L
0.3
0.2
0.2
0.4
0.6
L1 /L, L2 /L
(b)
2
2
0.0
0
For
obtaining
L2 /L
01
0
0.
4
0.0
6
8
0.0
eB /B 0.0
0.10
0.12
0.14
0.16
0.1
0.0
4
0.0
0.
0
0.8
1
1.0
For
obtaining
L1 /L
Figure 3.21 Effective area
for the case of eL>L , 0.5 and
0 , eB>B , 16 (After Highter
and Anders, 1985) (Highter,
W. H. and Anders, J. C. (1985).
“Dimensioning Footings
Subjected to Eccentric Loads,”
Journal of Geotechnical
Engineering, American Society
of Civil Engineers, Vol. 111,
No. GT5, pp. 659–665. With
permission from ASCE.)
The effective width is
Ar
L
(3.62)
Lr 5 L
(3.63)
Br 5
The effective length is
The magnitudes of B1 and B2 can be determined from Figure 3.22b.
Case IV. eL>L , 16 and eB>B , 16. Figure 3.23a shows the effective area for this case. The
ratio B2>B, and thus B2, can be determined by using the eL>L curves that slope upward.
Similarly, the ratio L2>L, and thus L2, can be determined by using the eL>L curves that
slope downward. The effective area is then
Ar 5 L2B 1 12 (B 1 B2 ) (L 2 L2 )
(3.64)
168 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
B1
eB
eL
Qult
L
Effective
area
B2
B
(a)
0.5
eL /L 0.4
0.167
0.1
0.08
0.06
eB /B
0.3
0.2
0.2
0.4
0.6
B1 /B, B2 /B
(b)
2
2
0.0
0
For
obtaining
B2 /B
01
0
0.
4
0.0
6
8
0.0
eL /L 0.0
0.10
0.12
0.14
0.16
0.1
0.0
4
0.0
0.
0
0.8
1
1.0
For
obtaining
B1 /B
Figure 3.22 Effective area for the case of eL>L , 16 and 0 , eB>B , 0.5 (After Highter
and Anders, 1985) Highter, W. H. and Anders, J. C. (1985). “Dimensioning Footings
Subjected to Eccentric Loads,” Journal of Geotechnical Engineering, American Society
of Civil Engineers, Vol. 111, No. GT5, pp. 659–665. With permission from ASCE.)
The effective width is
Ar
L
(3.65)
Lr 5 L
(3.66)
Br 5
The effective length is
Case V. (Circular Foundation) In the case of circular foundations under eccentric
loading (Figure 3.24a), the eccentricity is always one way. The effective area Ar and the
effective width Br for a circular foundation are given in a nondimensional form in Table 3.8.
Once A9 and B9 are determined, the effective length can be obtained as
Lr 5
Ar
Br
3.11 Bearing Capacity–Two-way Eccentricity
169
B
L2
eB
eL
L
Qult
Effective
area
B2
(a)
For obtaining B2 /B
0.16
0.14
0.12
0.20
0.10
0.15
0.08
eB /B
1
0.
0.06
0.0
0.10
0.1
4
8
0.04
0.0
6
0.05
0.04
0.02 eL /L
eL/L 0.02
For obtaining L2/L
0
0
0.2
0.4
0.6
B2 /B, L2 /L
(b)
0.8
eR
Qult
R
Figure 3.24 Effective area for circular foundation
1.0
Figure 3.23 Effective area for the
case of eL>L , 16 and eB>B , 16
(After Highter and Anders, 1985)
(Highter, W. H. and Anders, J. C.
(1985). “Dimensioning Footings
Subjected to Eccentric Loads,” Journal
of Geotechnical Engineering,
American Society of Civil Engineers,
Vol. 111, No. GT5, pp. 659–665. With
permission from ASCE.)
170 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
Table 3.8 Variation of Ar>R2 and Br>R with
eR>R for Circular Foundations
eR9 , R
A9 , R 2
B9 , R
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
2.8
2.4
2.0
1.61
1.23
0.93
0.62
0.35
0.12
0
1.85
1.32
1.2
0.80
0.67
0.50
0.37
0.23
0.12
0
Example 3.8
A square foundation is shown in Figure 3.25, with eL 5 0.3 m and eB 5 0.15 m. Assume two-way eccentricity, and determine the ultimate load, Qult.
Solution
We have
eL
0.3
5
5 0.2
L
1.5
and
eB
0.15
5
5 0.1
B
1.5
This case is similar to that shown in Figure 3.21a. From Figure 3.21b, for eL>L 5 0.2
and eB>B 5 0.1,
L1
< 0.85;
L
L1 5 (0.85) (1.5) 5 1.275 m
L2
< 0.21;
L
L2 5 (0.21) (1.5) 5 0.315 m
and
From Eq. (3.58),
Ar 5 12 (L1 1 L2 )B 5 12 (1.275 1 0.315) (1.5) 5 1.193 m2
3.11 Ultimate Bearing Capacity–Two-way Eccentricity
Sand
18 kN/m3
30
c 0
0.7 m
1.5 m 1.5 m
eB 0.15 m
1.5 m
eL 0.3 m
Figure 3.25 An eccentrically loaded
foundation
1.5 m
From Eq. (3.60),
Lr 5 L1 5 1.275 m
From Eq. (3.59),
Br 5
Ar
1.193
5
5 0.936 m
Lr
1.275
Note from Eq. (3.40) with cr 5 0,
qur 5 qNqFqsFqdFqi 1 12gBrNgFgsFgdFgi
where q 5 (0.7) (18) 5 12.6 kN>m2.
For fr 5 30°, from Table 3.3, Nq 5 18.4 and Ng 5 22.4. Thus from Table 3.4,
Fqs 5 1 1 ¢
Br
0.936
≤ tan fr 5 1 1 ¢
≤ tan 30° 5 1.424
Lr
1.275
Fgs 5 1 2 0.4 ¢
Br
0.936
≤ 5 0.706
≤ 5 1 2 0.4 ¢
Lr
1.275
Fqd 5 1 1 2 tan fr(1 2 sin fr) 2
Df
B
511
and
Fgd 5 1
(0.289) (0.7)
5 1.135
1.5
171
172 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
So
Qult 5 Arqur 5 Ar(qNqFqsFqd 1 12gBrNgFgsFgd )
5 (1.193) 3(12.6) (18.4) (1.424) (1.135)
1(0.5) (18) (0.936) (22.4) (0.706) (1)4 < 606 kN
■
Example 3.9
Consider the foundation shown in Figure 3.25 with the following changes:
eL ⫽ 0.18 m
eB ⫽ 0.12 m
For the soil, ␥ ⫽ 16.5 kN/m3
␾⬘ ⫽ 25°
c⬘ ⫽ 25 kN/m2
Determine the ultimate load, Qult.
Solution
eL
0.18
5
5 0.12;
L
1.5
eB
0.12
5
5 0.08
B
1.5
This is the case shown in Figure 3.23a. From Figure 3.23b,
B2
< 0.1;
B
L2
< 0.32
L
So
B2 ⫽ (0.1)(1.5) ⫽ 0.15 m
L2 ⫽ (0.32)(1.5) ⫽ 0.48 m
From Eq. (3.64),
1
1
Ar 5 L2B 1 (B 1 B2 ) (L 2 L2 ) 5 (0.48) (1.5) 1 (1.5 1 0.15) (1.5 2 0.48)
2
2
5 0.72 1 0.8415 5 1.5615 m2
Br 5
Ar
1.5615
5
5 1.041m
L
1.5
L9 5 1.5 m
From Eq. (3.40),
1
qur 5 crNcFcs Fed 1 qNqFqsFqd 1 gBrNgFgsFgd
2
3.12 Bearing Capacity of a Continuous Foundation Subjected to Eccentric Inclined Load
173
For ␾⬘ ⫽ 25°, Table 3.3 gives Nc ⫽ 20.72, Nq ⫽ 10.66 and N␥ ⫽ 10.88. From Table 3.4,
Fcs 5 1 1 a
Br Nq
1.041 10.66
ba b 5 1 1 a
ba
b 5 1.357
Lr Nc
1.5
20.72
Fqs 5 1 1 a
Br
1.041
b tanfr 5 1 1 a
b tan 25 5 1.324
Lr
1.5
Fgs 5 1 2 0.4 a
Br
1.041
b 5 120.4 a
b 5 0.722
Lr
1.5
Df
0.7
Fqd 5 1 1 2 tan fr(1 2 sinfr) 2 a b 5 1 1 2 tan 25(12sin 25) 2 a b 5 1.145
B
1.5
Fcd 5 Fqd 2
1 2 Fqd
9
Nc tan f
5 1.145 2
1 2 1.145
5 1.16
20.72 tan 25
Fgd 5 1
Hence,
qru 5 (25) (20.72) (1.357) (1.16) 1 (16.5 3 0.7) (10.66) (1.324) (1.145)
1
1 (16.5) (1.041) (10.88) (0.722) (1)
2
5 815.39 1 186.65 1 67.46 5 1069.5 kN>m2
Qult ⫽ A⬘qu⬘ ⫽ (1069.5)(1.5615) ⫽ 1670 kN
3.12
■
Bearing Capacity of a Continuous Foundation
Subjected to Eccentric Inclined Loading
The problem of ultimate bearing capacity of a continuous foundation subjected to an
eccentric inclined load was studied by Saran and Agarwal (1991). If a continuous foundation is located at a depth Df below the ground surface and is subjected to an eccentric load
(load eccentricity 5 e) inclined at an angle b to the vertical, the ultimate capacity can be
expressed as
1
(3.67)
Qult 5 B ccrNc(ei) 1 qNq(ei) 1 gBNg(ei) d
2
where Nc(ei), Nq(ei), and Ng(ei) 5 bearing capacity factors
q 5 gDf
The variations of the bearing capacity factors with e>B, fr, and b derived by Saran and
Agarwal are given in Figures 3.26, 3.27, and 3.28.
174 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
100
60
=0
80
50
= 10
40
Nc (ei)
Nc (ei)
60
e/B = 0
40
30
0.1
0.2
20
20
0.3
e/B = 0
0.1
0.2
10
0
0
10
20
30
Soil friction angle, (deg)
0.3
40
(a)
0
0
10
20
30
Soil friction angle, (deg)
40
(b)
40
= 20
30
30
20
e/B = 0
Nc(ei)
Nc (ei)
= 30
0.1
20
e/B = 0
0.2
10
10
0.3
0.1
0.2
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
40
(c)
Figure 3.26 Variation of Nc(ei) with fr, e>B, and b
0
10
20
30
Soil friction angle, (deg)
(d)
40
3.12 Bearing Capacity of a Continuous Foundation Subjected to Eccentric Inclined Load
175
50
80
= 10
40
60
=0
Nq (ei)
Nq (ei)
30
40
e/B = 0
20
e/B = 0
0.2
0.1
0.2
20
0.1
10
0.3
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
0
40
10
20
30
Soil friction angle, (deg)
40
(b)
(a)
30
30
= 20
= 30
e/B = 0
10
Nq (ei)
20
Nq(ei)
20
0.1
10
0.2
e/B = 0
0.3
0.1
0.2
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
0
40
10
20
30
Soil friction angle, (deg)
40
(d)
(c)
Figure 3.27 Variation of Nq(ei) with fr, e>B, and b
Example 3.10
A continuous foundation is shown in Figure 3.29. Estimate the ultimate load, Qult per
unit length of the foundation.
Solution
With cr 5 0, from Eq. (3.67),
1
Qult 5 cqNq(ei) 1 gBNg(ei) d
2
176 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
160
=0
120
= 10
80
e/B = 0
N (ei)
N (ei)
80
40
e/B = 0
40
0.1
0.1
0.2
0.2
0.3
0.3
0
0
0
10
20
30
Soil friction angle, (deg)
10
40
20
30
Soil friction angle, (deg)
40
(b)
(a)
30
60
= 20
20
40
N(ei)
N (ei)
= 30
e/B = 0
0.1
20
e/B = 0
0.1
10
0.2
0.3
0.2
0.3
0
0
20
30
Soil friction angle, (deg)
40
30
35
Soil friction angle, (deg)
40
(d)
(c)
Figure 3.28 Variation of Ng(ei) with fr, e>B, and b
B 5 1.5 m, q 5 Dfg 5 (1) (16) 5 16 kN>m2, e>B 5 0.15>1.5 5 0.1, and b 520°.
From Figures 3.27(c) and 3.28(c), Nq(ei) 5 14.2 and Ng(ei) 5 20. Hence,
Qult 5 (1.5) 3(16) (14.2) 1 ( 12 ) (16) (1.5) (20) 4 5 700.8 kN , m
Problems
177
Qult
20
16 kN/m3
35
c 0
1m
0.15 m
1.5 m
■
Figure 3.29
Problems
3.1
For the following cases, determine the allowable gross vertical load-bearing capacity
of the foundation. Use Terzaghi’s equation and assume general shear failure in soil.
Use FS 5 4.
Part
a.
b.
c.
3.2
3.3
3.4
3.5
3.6
B
1.22 m
2m
3m
Df
0.91 m
1m
2m
f9
25°
30°
30°
g
c9
2
28.75 kN>m
0
0
Foundation type
3
17.29 kN>m
17 kN>m3
16.5 kN>m3
Continuous
Continuous
Square
A square column foundation has to carry a gross allowable load of 1805 kN
(FS 5 3). Given: Df 5 1.5 m, g 5 15.9 kN>m3, fr 5 34°, and cr 5 0. Use
Terzaghi’s equation to determine the size of the foundation (B). Assume general
shear failure.
Use the general bearing capacity equation [Eq. (3.19)] to solve the following:
a. Problem 3.1a
b. Problem 3.1b
c. Problem 3.1c
The applied load on a shallow square foundation makes an angle of 15° with the vertical. Given: B 5 1.83 m, Df 5 0.9 m, g 5 18.08 kN>m3, fr 5 25°, and cr 5
23.96 kN>m2. Use FS 5 4 and determine the gross allowable load. Use Eq. (3.19).
A column foundation (Figure P3.5) is 3 m 3 2 m in plan. Given: Df 5 1.5 m,
fr 5 25°, cr 5 70 kN>m2. Using Eq. (3.19) and FS 5 3, determine the net
allowable load [see Eq. (3.15)] the foundation could carry.
For a square foundation that is B 3 B in plan, Df 5 2 m; vertical gross allowable
load, Qall 5 3330 kN, g 5 16.5 kN>m3; fr 5 30°; cr 5 0; and FS 5 4. Determine
the size of the foundation. Use Eq. (3.19).
178 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
17 kN/m3
1m
1.5 m
Groundwater level
sat 19.5 kN/m3
3m2m
Figure P3.5
3.7
For the design of a shallow foundation, given the following:
Soil:
fr 5 25°
cr 5 50 kN>m2
Unit weight, g 5 17 kN>m3
Modulus of elasticity, Es 5 1020 kN>m2
Poisson’s ratio, ms 5 0.35
Foundation:
L 5 1.5 m
B51m
Df 5 1 m
Calculate the ultimate bearing capacity. Use Eq. (3.27).
3.8 An eccentrically loaded foundation is shown in Figure P3.8. Use FS of 4 and determine the maximum allowable load that the foundation can carry. Use Meyerhof’s effective area method.
3.9 Repeat Problem 3.8 using Prakash and Saran’s method.
3.10 For an eccentrically loaded continuous foundation on sand, given B ⫽ 1.8 m, Df ⫽
0.9 m, e/B ⫽ 0.12 (one-way eccentricity), ␥ ⫽ 16 kN/m3, and ␾r ⫽ 35°. Using the
reduction factor method, estimate the ultimate load per unit length of the foundation.
(Eccentricity
in one direction
only) 0.1 m
0.8 m
Qall
17 kN/m3
c 0
32
1.5 m 1.5 m
Centerline
Figure P3.8
3.11 An eccentrically loaded continuous foundation is shown in Figure P3.11. Determine
the ultimate load Qu per unit length that the foundation can carry. Use the reduction
factor method.
3.12 A square footing is shown in Figure P3.12. Use FS 5 6, and determine the size of
the footing. Use Prakash and Saran theory [Eq. (3.43)].
References
179
Qu
16.5 kN/m3
Groundwater table
0.61 m
1.22 m
sat 18.55 kN/m3
c 0
35
0.61 m
1.52 m
Figure P3.11
450 kN
70 kN•m
16 kN/m3
c 0
30
1.2 m
BB
Water table
sat 19 kN/m3
c 0
30
Figure P3.12
3.13 The shallow foundation shown in Figure 3.19 measures 1.2 m 3 1.8 m and is subjected to a centric load and a moment. If eB 5 0.12 m, eL 5 0.36 m, and the depth of
the foundation is 1 m, determine the allowable load the foundation can carry. Use a
factor of safety of 3. For the soil, we are told that unit weight g 5 17 kN>m3, friction
angle fr 5 35°, and cohesion cr 5 0.
References
BOZOZUK, M. (1972). “Foundation Failure of the Vankleek Hill Tower Site,” Proceedings, Specialty Conference on Performance of Earth and Earth-Supported Structures, Vol. 1, Part 2,
pp. 885–902.
BRAND, E. W., MUKTABHANT, C., and TAECHANTHUMMARAK, A. (1972). “Load Test on Small Foundations in Soft Clay,” Proceedings, Specialty Conference on Performance of Earth and EarthSupported Structures, American Society of Civil Engineers, Vol. 1, Part 2, pp. 903–928.
CAQUOT, A., and KERISEL, J. (1953). “Sur le terme de surface dans le calcul des fondations en milieu
pulverulent,” Proceedings, Third International Conference on Soil Mechanics and Foundation
Engineering, Zürich, Vol. I, pp. 336–337.
DE BEER, E. E. (1970). “Experimental Determination of the Shape Factors and Bearing Capacity
Factors of Sand,” Geotechnique, Vol. 20, No. 4, pp. 387– 411.
HANNA, A. M., and MEYERHOF, G. G. (1981). “Experimental Evaluation of Bearing Capacity of Footings
Subjected to Inclined Loads,” Canadian Geotechnical Journal, Vol. 18, No. 4, pp. 599–603.
180 Chapter 3: Shallow Foundations: Ultimate Bearing Capacity
HANSEN, J. B. (1970). A Revised and Extended Formula for Bearing Capacity, Bulletin 28, Danish
Geotechnical Institute, Copenhagen.
HIGHTER, W. H., and ANDERS, J. C. (1985). “Dimensioning Footings Subjected to Eccentric Loads,”
Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 111, No.
GT5, pp. 659–665.
KUMBHOJKAR, A. S. (1993). “Numerical Evaluation of Terzaghi’s Ng,” Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 119, No. 3, pp. 598–607.
MEYERHOF, G. G. (1953). “The Bearing Capacity of Foundations Under Eccentric and Inclined
Loads,” Proceedings, Third International Conference on Soil Mechanics and Foundation Engineering, Zürich, Vol. 1, pp. 440–445.
MEYERHOF, G. G. (1963). “Some Recent Research on the Bearing Capacity of Foundations,” Canadian Geotechnical Journal, Vol. 1, No. 1, pp. 16–26.
PRAKASH, S., and SARAN, S. (1971). “Bearing Capacity of Eccentrically Loaded Footings,” Journal
of the Soil Mechanics and Foundations Division, ASCE, Vol. 97, No. SM1, pp. 95–117.
PRANDTL, L. (1921). “Über die Eindringungsfestigkeit (Härte) plastischer Baustoffe und die Festigkeit von Schneiden,” Zeitschrift für angewandte Mathematik und Mechanik, Vol. 1, No. 1,
pp. 15–20.
PURKAYASTHA, R.D., and CHAR, R. A. N. (1977). “Stability Analysis of Eccentrically Loaded Footings,” Journal of Geotechnical Engineering Div., ASCE, Vol. 103, No. 6, pp. 647–651.
REISSNER, H. (1924). “Zum Erddruckproblem,” Proceedings, First International Congress of Applied
Mechanics, Delft, pp. 295–311.
SARAN, S., and AGARWAL, R. B. (1991). “Bearing Capacity of Eccentrically Obliquely Loaded Footing,” Journal of Geotechnical Engineering, ASCE, Vol. 117, No. 11, pp. 1669–1690.
TERZAGHI, K. (1943). Theoretical Soil Mechanics, Wiley, New York.
VESIC, A. S. (1963). “Bearing Capacity of Deep Foundations in Sand,” Highway Research Record
No. 39, National Academy of Sciences, pp. 112–153.
VESIC, A. S. (1973). “Analysis of Ultimate Loads of Shallow Foundations,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 99, No. SM1,
pp. 45–73.
4
4.1
Ultimate Bearing Capacity of Shallow
Foundations: Special Cases
Introduction
The ultimate bearing capacity problems described in Chapter 3 assume that the soil supporting the foundation is homogeneous and extends to a great depth below the bottom of
the foundation. They also assume that the ground surface is horizontal. However, that is
not true in all cases: It is possible to encounter a rigid layer at a shallow depth, or the soil
may be layered and have different shear strength parameters. In some instances, it may be
necessary to construct foundations on or near a slope, or it may be required to design a
foundation subjected to uplifting load.
This chapter discusses bearing capacity problems relating to these special cases.
4.2
Foundation Supported by a Soil with a Rigid Base
at Shallow Depth
Figure 4.1(a) shows a shallow, rough continuous foundation supported by a soil that
extends to a great depth. Neglecting the depth factor, for vertical loading Eq. (3.19) will
take the form
qu 5 crNc 1 qNq 1
1
gBNg
2
(4.1)
The general approach for obtaining expressions for Nc, Nq, and Ng was outlined in
Chapter 3. The extent of the failure zone in soil, D, at ultimate load obtained in the
derivation of Nc and Nq by Prandtl (1921) and Reissner (1924) is given in Figure 4.1(b).
Similarly, the magnitude of D obtained by Lundgren and Mortensen (1953) in evaluating Ng is given in the figure.
Now, if a rigid, rough base is located at a depth of H , D below the bottom of the foundation, full development of the failure surface in soil will be restricted. In such a case, the soil
failure zone and the development of slip lines at ultimate load will be as shown in Figure 4.2.
181
182 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
B
qu
q = Df
Df
45 – /2
45 + /2
c
D
(a)
3
Nc and
Nq
D/B
2
N
1
0
0
10
20
30
40
Soil friction angle, (deg)
50
(b)
Figure 4.1 (a) Failure surface under a rough continuous foundation;
(b) variation of D>B with soil friction angle fr
B
qu
q = Df
H
c
Figure 4.2 Failure surface under a rough, continuous foundation with a rigid, rough base
located at a shallow depth
4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth
183
Mandel and Salencon (1972) determined the bearing capacity factors applicable to this case by
numerical integration, using the theory of plasticity. According to their theory, the ultimate
bearing capacity of a rough continuous foundation with a rigid, rough base located at a shallow
depth can be given by the relation
qu 5 crN *c 1 qN *q 1
1
gBN *g
2
(4.2)
where
N *c , N *q , N *g 5 modified bearing capacity factors
B 5 width of foundation
g 5 unit weight of soil
Note that, for H > D, N*c 5 Nc , N*q 5 Nq , and N*g 5 Ng (Lundgren and Mortensen,
1953). The variations of N*c , N*q , and N*g with H>B and the soil friction angle fr are given
in Figures 4.3, 4.4, and 4.5, respectively.
10,000
5000
H/B = 0.25
2000
0.33
1000
500
0.5
200
N *c
100
D/B =
2.4
1.0
50
1.6
20
1.2
10
5
0.9
0.7
2
1
0
10
20
(deg)
30
40
Figure 4.3 Mandel and
Salencon’s bearing capacity
factor N *c [Eq. (4.2)]
184 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
10,000
5000
2000
1000
0.4
H/B = 0.2
500
0.6
200
N *q
D/B =
3.0
1.0
100
2.4
50
1.9
20
1.6
10
5
1.4
1.2
2
1
20
25
30
35
(deg)
40
45
Figure 4.4 Mandel and
Salencon’s bearing capacity
factor N*q [Eq. (4.2)]
Rectangular Foundation on Granular Soil
Neglecting the depth factors, the ultimate bearing capacity of rough circular and rectangular foundations on a sand layer (cr 5 0) with a rough, rigid base located at a shallow
depth can be given as
qu 5 qN*q F*qs 1
1
gBN*g F*gs
2
(4.3)
*
*
, Fgs
5 modified shape factors.
where Fqs
The shape factors F*qs and F*gs are functions of H>B and fr. On the basis of the
work of Meyerhof and Chaplin (1953), and simplifying the assumption that, in radial
planes, the stresses and shear zones are identical to those in transverse planes, Meyerhof
(1974) proposed that
4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth
185
10,000
5000
2000
1000
H/B = 0.2
500
0.4
200
0.6
N *
D/B =
1.5
1.0
100
1.2
50
1.0
20
0.8
10
0.6
5
2
0.5
Figure 4.5 Mandel and
Salencon’s bearing capacity
factor Ng* [Eq. (4.2)]
1
20
25
30
35
(deg)
40
45
F*qs < 1 2 m1 ¢
B
≤
L
(4.4)
F*gs < 1 2 m2 ¢
B
≤
L
(4.5)
and
where L 5 length of the foundation. The variations of m1 and m2 with H>B and fr are
shown in Figure 4.6.
More recently, Cerato and Lutenegger (2006) provided some test results for the bearing capacity factor, N*. These tests were conducted using square and circular plates with B
varying from 0.152 m (6 in.) to 0.305 m (12 in.). It was assumed that Terzaghi’s bearingcapacity equations for square and circular foundations can be used. Or, from Eqs. (3.10)
and (3.11) with c 0,
qu qNq* 0.4BN* (square foundation)
(4.6)
186 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
1.0
H/B = 0.1
0.8
0.2
0.6
m1
0.4
0.4
0.6
1.0
0.2
2.0
0
20
25
35
30
(deg)
40
45
1.0
H/B = 0.1
0.8
0.2
0.4
0.6
m2
0.6
1.0
0.4
0.2
0
20
25
35
30
(deg)
40
45
Figure 4.6 Variation of m1 and m2
with H>B and fr
and
qu qNq* 0.3BN* (circular foundation)
(4.7)
The experimentally determined variation of N* is shown in Figure 4.7. It also was observed in
this study that N* becomes equal to N at H/B ⬇ 3 instead of D/B, as shown in Figure 4.5. For
that reason, Figure 4.7 shows the variation of N* for H/B 0.5 to 3.0.
Foundation on Saturated Clay
For saturated clay (i.e., under the undrained condition, or f 5 0), Eq. (4.2) will simplify
to the form
qu 5 cuN *c 1 q
(4.8)
4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth
187
2000
1500
N *γ 1000
H/B = 0.5
1.0
500
2.0
3.0
0
20
25
35
30
Friction angle, φ (deg)
40
45
Figure 4.7 Cerato and Lutenegger’s
test results for N*
Mandel and Salencon (1972) performed calculations to evaluate N *c for continuous foundations. Similarly, Buisman (1940) gave the following relationship for obtaining the ultimate
bearing capacity of square foundations:
qu(square) 5 ¢ p 1 2 1
B
"2
2
≤ cu 1 q
2H
2
¢ for
B
"2
2
> 0≤
2H
2
(4.9)
In this equation, cu is the undrained shear strength.
Equation (4.9) can be rewritten as
B
2 0.707
H
5 5.14 £1 1
≥ cu 1 q
5.14
(''''')'''
''*
0.5
qu(square)
N*c(square)
Table 4.1 gives the values of N *c for continuous and square foundations.
(4.10)
188 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
Table 4.1 Values of N c* for Continuous and
Square Foundations (f 5 0)
N *c
B
H
2
3
4
5
6
8
10
a
b
Squarea
Continuousb
5.43
5.93
6.44
6.94
7.43
8.43
9.43
5.24
5.71
6.22
6.68
7.20
8.17
9.05
Buisman’s analysis (1940)
Mandel and Salencon’s analysis (1972)
Example 4.1
A square foundation measuring 0.76 m 3 0.76 m is constructed on a layer of sand. We
are given that Df 5 0.61 m, g 5 17.29 kN>m3, fr 5 35°, and cr 5 0. A rock layer is
located at a depth of 0.46 m below the bottom of the foundation. Using a factor of safety
of 4, determine the gross allowable load the foundation can carry.
Solution
From Eq. (4.3),
qu 5 qN *qF *qs 1
1
gBN *gF *gs
2
and we also have
q 5 17.29 3 0.61 5 10.55 kN>m2
For fr 5 35°, H>B 5 0.46>0.76 m 5 0.6, N *q < 90 (Figure 4.4), and N *g < 50
(Figure 4.5), and we have
F *qs 5 1 2 m1 (B>L)
From Figure 4.6(a), for fr 5 35°, H>B 5 0.6, and the value of m1 5 0.34, so
F *qs 5 1 2 (0.34) (0.76>0.76) 5 0.66
Similarly,
F *gs 5 1 2 m2 (B>L)
From Figure 4.6(b), m2 5 0.45, so
F *gs 5 1 2 (0.45) (0.76>0.76) 5 0.55
4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth
189
Hence,
qu 5 (10.55) (90) (0.66) 1 (1>2) (17.29) (0.76) (50) (0.55) 5 807.35 kN>m2
and
Qall 5
quB2
(807.35) (0.76 3 0.76)
5
5 116.58 kN
FS
4
■
Example 4.2
Solve Example 4.1 using Eq. (4.6).
Solution
From Eq. (4.6),
qu qNq* 0.4BN*
For 35° and H/B 0.6, the value of Nq* ⬇ 90 (Figure 4.4) and N* ⬇ 230
(Figure 4.7).
So
qu (10.55)(90) (0.4)(17.29)(0.76)(230) 949.5 1208.9 2158.4 kN/m2
Q all 5
quB2
< 311.7 kN
FS
■
Example 4.3
Consider a square foundation 1 m 3 1 m in plan located on a saturated clay layer underlain by a layer of rock. Given:
Clay: cu 5 72 kN>m2
Unit weight: g 5 18 kN>m3
Distance between the bottom of foundation and the rock layer 5 0.25 m
Df 5 1 m
Estimate the gross allowable bearing capacity of the foundation. Use FS 3.
Solution
From Eq. (4.10),
0.5
qu 5 5.14 £1 1
B
2 0.707
H
≥ cu 1 q
5.14
190 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
For B>H 5 1>0.25 5 4; cu 5 72 kN>m2; and q 5 g Df 5 (18) (1) 5 18 kN>m3.
qu 5 5.14B1 1
(0.5) (4) 2 0.707
R72 1 18 5 481.2 kN>m2
5.14
qall 5
4.3
qu
481.2
5
5 160.4 kN , m2
FS
3
■
Bearing Capacity of Layered Soils: Stronger Soil
Underlain by Weaker Soil
The bearing capacity equations presented in Chapter 3 involve cases in which the soil
supporting the foundation is homogeneous and extends to a considerable depth. The
cohesion, angle of friction, and unit weight of soil were assumed to remain constant for
the bearing capacity analysis. However, in practice, layered soil profiles are often
encountered. In such instances, the failure surface at ultimate load may extend through
two or more soil layers, and a determination of the ultimate bearing capacity in layered
soils can be made in only a limited number of cases. This section features the procedure
for estimating the bearing capacity for layered soils proposed by Meyerhof and Hanna
(1978) and Meyerhof (1974).
Figure 4.8 shows a shallow, continuous foundation supported by a stronger soil
layer, underlain by a weaker soil that extends to a great depth. For the two soil layers, the
physical parameters are as follows:
Soil properties
Layer
Top
Bottom
Unit weight
Friction
angle
Cohesion
g1
g2
f1r
f2r
c1r
c2r
At ultimate load per unit area (qu ), the failure surface in soil will be as shown in the
figure. If the depth H is relatively small compared with the foundation width B, a punching shear failure will occur in the top soil layer, followed by a general shear failure in the
bottom soil layer. This is shown in Figure 4.8(a). However, if the depth H is relatively
large, then the failure surface will be completely located in the top soil layer, which is the
upper limit for the ultimate bearing capacity. This is shown in Figure 4.8b.
The ultimate bearing capacity for this problem, as shown in Figure 4.8a, can be
given as
qu 5 qb 1
2(Ca 1 Pp sin dr)
B
2 g1H
(4.11)
4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain by Weaker Soil
191
B
qu
Df
a
b
Ca
H
Ca
Pp
Pp
a
Stronger soil
1
1
c1
b
(a)
B
Weaker soil
2
2
c2
qu
Df
Stronger soil
1
1
c1
H
(b)
Weaker soil
2
2
c2
Figure 4.8 Bearing capacity of a
continuous foundation on layered
soil
where
B 5 width of the foundation
Ca 5 adhesive force
Pp 5 passive force per unit length of the faces aar and bbr
qb 5 bearing capacity of the bottom soil layer
dr 5 inclination of the passive force Pp with the horizontal
Note that, in Eq. (4.11),
Ca 5 car H
where car 5 adhesion.
Equation (4.11) can be simplified to the form
qu 5 qb 1
2Df KpH tan dr
2car H
1 g1H2 ¢1 1
≤
2 g1H
B
H
B
(4.12)
where KpH 5 horizontal component of passive earth pressure coefficient.
However, let
KpH tan dr 5 Ks tan f1r
(4.13)
192 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
where Ks 5 punching shear coefficient. Then,
qu 5 qb 1
2Df Ks tan f1r
2car H
1 g1H2 ¢ 1 1
≤
2 g1H
B
H
B
(4.14)
The punching shear coefficient, Ks , is a function of q2>q1 and fr1 , or, specifically,
Ks 5 f¢
q2
, fr ≤
q1 1
Note that q1 and q2 are the ultimate bearing capacities of a continuous foundation of
width B under vertical load on the surfaces of homogeneous thick beds of upper and lower
soil, or
q1 5 c1r Nc(1) 1 12g1BNg(1)
(4.15)
q2 5 c2r Nc(2) 1 12g2BNg(2)
(4.16)
and
where
Nc(1) , Ng(1) 5 bearing capacity factors for friction angle f1r (Table 3.3)
Nc(2) , Ng(2) 5 bearing capacity factors for friction angle f2r (Table 3.3)
Observe that, for the top layer to be a stronger soil, q2 >q1 should be less than unity.
The variation of Ks with q2>q1 and f1r is shown in Figure 4.9. The variation of
car >c1r with q2>q1 is shown in Figure 4.10. If the height H is relatively large, then the
40
30
KS
q2
q1 = 1
20
0.4
10
0.2
0
0
20
30
40
1 (deg)
50
Figure 4.9 Meyerhof and Hanna’s
punching shear coefficient Ks
4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain by Weaker Soil
193
1.0
0.9
ca
0.8
c1
0.7
0.6
0
0.2
0.4
0.6
0.8
1.0
q2
q1
Figure 4.10 Variation
of car >c1r with q2>q1
based on the theory of
Meyerhof and Hanna
(1978)
failure surface in soil will be completely located in the stronger upper-soil layer
(Figure 4.8b). For this case,
qu 5 qt 5 c1r Nc(1) 1 qNq(1) 1 12 g1BNg(1).
(4.17)
where Nc(1), Nq(1), and Ng(1) 5 bearing capacity factors for fr 5 f1r (Table 3.3) and
q 5 g1Df.
Combining Eqs. (4.14) and (4.17) yields
qu 5 qb 1
2Df Ks tan f1r
2car H
1 g1H2 ¢1 1
≤
2 g1H < qt
B
H
B
(4.18)
For rectangular foundations, the preceding equation can be extended to the form
qu 5 qb 1 ¢1 1
B
L
≤¢
1 g1H2 ¢1 1
2car H
B
B
L
≤
≤ ¢1 1
2Df
H
≤¢
Ks tan f1r
B
(4.19)
≤ 2 g1H < qt
where
qb 5 c2r Nc(2)Fcs(2) 1 g1 (Df 1 H)Nq(2)Fqs(2) 1
1
g BNg(2)Fgs(2)
2 2
(4.20)
194 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
and
qt 5 c1r Nc(1)Fcs(1) 1 g1DfNq(1)Fqs(1) 1
1
g BNg(1)Fgs(1)
2 1
(4.21)
in which
Fcs(1) , Fqs(1) , Fgs(1) 5 shape factors with respect to top soil layer (Table 3.4)
Fcs(2) , Fqs(2) , Fgs(2) 5 shape factors with respect to bottom soil layer (Table 3.4)
Special Cases
1. Top layer is strong sand and bottom layer is saturated soft clay (f2 5 0). From
Eqs. (4.19), (4.20), and (4.21),
qb 5 ¢1 1 0.2
B
≤ 5.14c2 1 g1 (Df 1 H)
L
(4.22)
and
qt 5 g1DfNq(1)Fqs(1) 1 12g1BNg(1)Fgs(1)
(4.23)
2Df Ks tan f1r
B
B
≤ 5.14c2 1 g1H2 ¢1 1 ≤ ¢ 1 1
≤
L
L
H
B
1
1 g1Df < g1DfNq(1)Fqs(1) 1 g1BNg(1)Fgs(1)
2
(4.24)
Hence,
qu 5 ¢ 1 1 0.2
where c2 5 undrained cohesion.
For a determination of Ks from Figure 4.9,
c2Nc(2)
q2
5.14c2
51
5
q1
0.5g1BNg(1)
2 g1BNg(1)
(4.25)
2. Top layer is stronger sand and bottom layer is weaker sand (cr1 5 0, cr2 5 0). The
ultimate bearing capacity can be given as
qu 5 Bg1 (Df 1 H)Nq(2)Fqs(2) 1
1
2
g2BNg(2)Fgs(2) R
2Df Ks tan f1r
1 g1H2 ¢1 1 ≤ ¢1 1
≤
2 g1H < qt
L
H
B
B
(4.26)
4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain by Weaker Soil
195
where
qt 5 g1DfNq(1)Fqs(1) 1
1
g BN F
2 1 g(1) gs(1)
(4.27)
Then
1
g2Ng(2)
q2
2 g2BNg(2)
51
5
q1
g1Ng(1)
2 g1BNg(1)
(4.28)
3. Top layer is stronger saturated clay (f1 5 0) and bottom layer is weaker saturated
clay (f2 5 0). The ultimate bearing capacity can be given as
qu 5 ¢1 1 0.2
B
B 2caH
≤5.14c2 1 ¢ 1 1 ≤ ¢
≤ 1 g1Df < qt
L
L
B
(4.29)
B
≤5.14c1 1 g1Df
L
(4.30)
where
qt 5 ¢1 1 0.2
and c1 and c2 are undrained cohesions. For this case,
5.14c2
c2
q2
5
5
q1
c1
5.14c1
(4.31)
Example 4.4
Refer to Figure 4.8(a) and consider the case of a continuous foundation with B 5 2 m,
Df 5 1.2 m , and H 5 1.5 m. The following are given for the two soil layers:
Top sand layer:
Unit weight g1 5 17.5 kN>m3
f1r 5 40°
c1r 5 0
Bottom clay layer:
Unit weight g2 5 16.5 kN>m3
fr2 5 0
c2 5 30 kN>m2
Determine the gross ultimate load per unit length of the foundation.
196 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
Solution
For this case, Eqs. (4.24) and (4.25) apply. For fr1 5 40°, from Table 3.3, Ng 5 109.41 and
c2Nc(2)
q2
(30) (5.14)
5
5
5 0.081
q1
0.5g1BNg(1)
(0.5)(17.5)(2)(109.41)
From Figure 4.9, for c2Nc(2)>0.5g1BNg(1) 5 0.081 and f1r 5 40°, the value of
Ks < 2.5 . Equation (4.24) then gives
qu 5 B1 1 (0.2) ¢
2Df
tan f1r
B
B
≤ R 5.14c2 1 ¢ 1 1 ≤g1H 2 ¢1 1
≤ Ks
1 g1Df
L
L
H
B
5 31 1 (0.2) (0) 4 (5.14) (30) 1 (1 1 0) (17.5) (1.5) 2
3 B1 1
(2) (1.2)
tan 40
R (2.5)
1 (17.5) (1.2)
1.5
2.0
5 154.2 1 107.4 1 21 5 282.6 kN>m2
Again, from Eq. (4.27),
qt 5 g1DfNq(1)Fqs(1) 1 12 g1BNg(1)Fgs(1)
From Table 3.3, for fr1 5 40°, Ng 5 109.4 and Nq 5 64.20 .
From Table 3.4,
Fqs(1) 5 1 1 ¢
B
≤ tan f1r 5 1 1 (0)tan 40 5 1
L
and
Fgs(1) 5 1 2 0.4
B
5 1 2 (0.4) (0) 5 1
L
so that
qt 5 (17.5) (1.2) (64.20) (1) 1 ( 12 ) (17.5) (2) (109.4) (1) 5 3262.7 kN>m2
Hence,
qu 5 282.6 kN>m2
Qu 5 (282.6) (B) 5 (282.6) (2) 5 565.2 kN , m
■
4.3 Bearing Capacity of Layered Soils: Stronger Soil Underlain by Weaker Soil
197
Example 4.5
A foundation 1.5 m 3 1 m is located at a depth Df of 1 m in a stronger clay. A softer
clay layer is located at a depth H of 3 ft, measured from the bottom of the foundation.
For the top clay layer,
Undrained shear strength 5 120 kN>m2
Unit weight 5 16.8 kN>m3
and for the bottom clay layer,
Undrained shear strength 5 48 kN>m2
Unit weight 5 16.2 kN>m3
Determine the gross allowable load for the foundation with an FS of 3.
Solution
For this problem, Eqs. (4.29), (4.30), and (4.31) will apply, or
qu 5 ¢1 1 0.2
B
B 2caH
≤5.14c2 1 ¢ 1 1 ≤ ¢
≤ 1 g1Df
L
L
B
< ¢1 1 0.2
B
≤ 5.14c1 1 g1Df
L
We are given the following data:
B51m
H51m
L 5 1.5 m
Df 5 1 m
g1 5 16.8 kN>m3
From Figure 4.10 for c2>c1 5 48>120 5 0.4, the value of ca>c1 < 0.9, so
ca 5 (0.9) (2500) 5 108 kN>m2
and
qu 5 B1 1 (0.2) ¢
(2) (108) (1)
1
1
≤ R (5.14) (48) 1 ¢ 1 1
≤B
R 1 (16.8) (1)
1.5
1.5
1
5 279.6 1 360 1 16.8 5 656.4 kN>m2
As a check, we have, from Eq. (4.30),
qt 5 B1 1 (0.2) ¢
1
≤ R (5.14) (120) 1 (16.8) (1)
1.5
5 699 1 16.8 5 715.8 kN>m2
Thus, qu 5 656.4 kN>m2 (i.e., the smaller of the two values just calculated), and
qall 5
qu
656.4
5
5 218.8 kN>m2
FS
3
The total allowable load is therefore
(qall ) (1 3 1.5) 5 328.2 kN
■
198 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
4.4
Bearing Capacity of Layered Soil: Weaker Soil
Underlain by Stronger Soil
When a foundation is supported by a weaker soil layer underlain by a stronger layer
(Figure 4.11a), the ratio of q2/q1 defined by Eqs. (4.15) and (4.16) will be greater than one.
Also, if H/B is relatively small, as shown in the left-hand half of Figure 4.11a, the failure
surface in soil at ultimate load will pass through both soil layers. However, for larger H/B
ratios, the failure surface will be fully located in the top, weaker soil layer, as shown in the
Weaker soil
γ1
φ ′1
Df
c′1
B
H
D
H
Stronger soil
γ2
φ ′2
c′2
Stronger soil
γ2
φ ′2
(a)
c′2
qu
qb
qt
D/B
(b)
H/B
Figure 4.11 (a) Foundation on weaker soil layer underlain by stronger sand layer, (b) Nature
of variation of qu with H/B
199
4.4 Bearing Capacity of Layered Soil: Weaker Soil Underlain by Stronger Soil
right-hand half of Figure 4.11a. For this condition, the ultimate bearing capacity
(Meyerhof, 1974; Meyerhof and Hanna, 1978) can be given by the empirical equation
qu 5 qt 1 (qb 2 qt ) a
H 2
b $ qt
D
(4.32)
where
D depth of failure surface beneath the foundation in the thick bed of the upper
weaker soil layer
qt ultimate bearing capacity in a thick bed of the upper soil layer
qb ultimate bearing capacity in a thick bed of the lower soil layer
So
1
qt 5 c1Nc(1)Fcs(1) 1 g1DfNq(1)Fqs(1) 1 g1BNg(1)Fgs(1)
2
(4.33)
1
qt 5 c2Nc(2)Fcs(2) 1 g2DfNq(2)Fqs(2) 1 g2BNg(2)Fgs(2)
2
(4.34)
and
where
Nc(1), Nq(1), N(1) bearing capacity factors corresponding to the soil friction angle 1
Nc(2), Nq(2), N(2) bearing capacity factors corresponding to the soil friction angle 2
Fcs(1), Fqs(1), Fs(1) shape factors corresponding to the soil friction angle 1
Fcs(2), Fqs(2), Fs(2) shape factors corresponding to the soil friction angle 2
Meyerhof and Hanna (1978) suggested that
•
•
D ⬇ B for loose sand and clay
D ⬇ 2B for dense sand
Equations (4.32), (4.33), and (4.34) imply that the maximum and minimum values of qu
will be qb and qt, respectively, as shown in Figure 4.11b.
Example 4.6
Refer to Figure 4.11a. For a layered saturated-clay profile, given: L 1.83 m, B 1.22 m,
Df 0.91 m, H 0.61 m, 1 17.29 kN/m3, 1 0, c1 57.5 kN/m2, 2 19.65 kN/m3,
2 0, and c2 119.79 kN/m2. Determine the ultimate bearing capacity of the foundation.
Solution
From Eqs. (4.15) and (4.16),
q2
c2Nc
c2
119.79
5
5 5
5 2.08 . 1
q1
c1
c1Nc
57.5
So, Eq. (4.32) will apply.
From Eqs. (4.33) and (4.34) with 1 2 0,
B
qt 5 a1 1 0.2 bNcc1 1 g1Df
L
5 c1 1 (0.2) a
1.22
b d (5.14) (57.5) 1 (0.91) (17.29) 5 334.96 1 15.73
1.83
5 350.69 kN>m2
200 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
and
qb 5 a1 1 0.2
B
bN c 1 g2Df
L c 2
5 c1 1 (0.2) a
1.22
b d (5.14) (119.79) 1 (0.91) (19.65)
1.83
5 697.82 1 17.88 5 715.7 kN>m2
From Eq. (4.32),
qu 5 qt 1 (qb 2 qt ) a
H 2
b
D
D<B
qu 5 350.69 1 (715.7 2 350.69) a
0.61 2
b < 442 kN>m2 . qt
1.22
Hence,
qu 442 kN/m2
4.5
■
Closely Spaced Foundations—Effect on Ultimate
Bearing Capacity
In Chapter 3, theories relating to the ultimate bearing capacity of single rough continuous
foundations supported by a homogeneous soil extending to a great depth were discussed.
However, if foundations are placed close to each other with similar soil conditions, the ultimate bearing capacity of each foundation may change due to the interference effect of the
failure surface in the soil. This was theoretically investigated by Stuart (1962) for granular soils. It was assumed that the geometry of the rupture surface in the soil mass would
be the same as that assumed by Terzaghi (Figure 3.5). According to Stuart, the following
conditions may arise (Figure 4.12).
Case I. (Figure 4.12a) If the center-to-center spacing of the two foundations is x $ x1,
the rupture surface in the soil under each foundation will not overlap. So the ultimate bearing capacity of each continuous foundation can be given by Terzaghi’s equation [Eq.
(3.3)]. For (cr 5 0)
qu 5 qNq 1 12 gBNg
(4.35)
where Nq, Ng 5 Terzaghi’s bearing capacity factors (Table 3.1).
Case II. (Figure 4.12b) If the center-to-center spacing of the two foundations
(x 5 x2 , x1 ) are such that the Rankine passive zones just overlap, then the magnitude of
qu will still be given by Eq. (4.35). However, the foundation settlement at ultimate load
will change (compared to the case of an isolated foundation).
Case III. (Figure 4.12c) This is the case where the center-to-center spacing of the two
continuous foundations is x 5 x3 , x2. Note that the triangular wedges in the soil
4.5 Closely Spaced Foundations—Effect on Ultimate Bearing Capacity
201
x x1
B
qu
2 2
B
qu
2 2
1
2 2
q Df
2 2
1
(a)
x x2
B
qu
B
qu
2 2
2 2
1
1
q Df
2 2
(b)
x x3
2
g1
B
qu
B
qu
3
3
e
d1
d2
(c)
q Df
2
g2
x x4
B
qu
B
qu
q Df
(d)
Figure 4.12 Assumptions for the failure surface in granular soil under two closely spaced rough
continuous foundations
(Note: a1 5 fr, a2 5 45 2 fr> 2, a3 5 180 2 2fr)
under the foundations make angles of 180° 2 2fr at points d1 and d2. The arcs of the
logarithmic spirals d1g1 and d1e are tangent to each other at d1. Similarly, the arcs of
the logarithmic spirals d2g2 and d2e are tangent to each other at d2. For this case, the
ultimate bearing capacity of each foundation can be given as (cr 5 0)
qu 5 qNqzq 1 21 gBNg zg
where zq, zg 5 efficiency ratios.
(4.36)
202 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
2.0
Rough base
Along this line two footings act as one
= 40
q 1.5
37
39
35
32
30
1.0
1
2
4
3
x/B
(a)
5
3.5
Rough base
Along this line two
footings act as one
3.0
2.5
= 40
39
2.0
37
35
32
1.5
30
1.0
1
2
3
x/B
(b)
4
5
Figure 4.13 Variation of efficiency ratios with x>B and fr
The efficiency ratios are functions of x>B and soil friction angle fr. The theoretical
variations of zq and zg are given in Figure 4.13.
Case IV. (Figure 4.12d): If the spacing of the foundation is further reduced such that
x 5 x4 , x3, blocking will occur and the pair of foundations will act as a single foundation. The soil between the individual units will form an inverted arch which travels down
with the foundation as the load is applied. When the two foundations touch, the zone of
203
4.6 Bearing Capacity of Foundations on Top of a Slope
arching disappears and the system behaves as a single foundation with a width equal to 2B.
The ultimate bearing capacity for this case can be given by Eq. (4.35), with B being
replaced by 2B in the second term.
The ultimate bearing capacity of two continuous foundations spaced close to each
other may increase since the efficiency ratios are greater than one. However, when the
closely spaced foundations are subjected to a similar load per unit area, the settlement Se
will be larger when compared to that for an isolated foundation.
4.6
Bearing Capacity of Foundations on Top of a Slope
In some instances, shallow foundations need to be constructed on top of a slope. In
Figure 4.14, the height of the slope is H, and the slope makes an angle b with the horizontal. The edge of the foundation is located at a distance b from the top of the slope.
At ultimate load, qu , the failure surface will be as shown in the figure.
Meyerhof (1957) developed the following theoretical relation for the ultimate bearing capacity for continuous foundations:
qu 5 crNcq 1 12 gBNgq
(4.37)
For purely granular soil, cr 5 0, thus,
qu 5 12 gBNgq
(4.38)
Again, for purely cohesive soil, f 5 0 (the undrained condition); hence,
qu 5 cNcq
(4.39)
where c 5 undrained cohesion.
b
B
qu
Df
H
c
Figure 4.14 Shallow foundation on top of a slope
204 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
400
Df
B
300
Df
=0
B
=1
0
200
Nq
20
= 40
40
= 40
100
0
20
0
50
= 30
30
40
25
= 30
0
10
5
30
1
0
1
2
3
b
B
4
5
6
Figure 4.15 Meyerhof’s
bearing capacity factor Ngq
for granular soil (cr 5 0)
The variations of Ngq and Ncq defined by Eqs. (4.38) and (4.39) are shown in
Figures 4.15 and 4.16, respectively. In using Ncq in Eq. (4.39) as given in Figure 4.16,
the following points need to be kept in mind:
1. The term
Ns 5
gH
c
(4.40)
is defined as the stability number.
2. If B , H, use the curves for Ns 5 0.
3. If B > H, use the curves for the calculated stability number Ns .
Stress Characteristics Solution for Granular Soil Slopes
For slopes in granular soils, the ultimate bearing capacity of a continuous foundation can
be given by Eq. (4.38), or
1
qu 5 gBNgq
2
On the basis of the method of stress characteristics, Graham, Andrews, and Shields (1988)
provided a solution for the bearing capacity factor N q for a shallow continuous foundation
on the top of a slope in granular soil. Figure 4.17 shows the schematics of the failure zone
in the soil for embedment (Df/B) and setback (b/B) assumed for those authors’ analysis. The
variations of N q obtained by this method are shown in Figures 4.18, 4.19, and 4.20.
4.6 Bearing Capacity of Foundations on Top of a Slope
8
6
Df
0
B
Ns 0
Df
1
B
0
15
30
45
60
90
Ns 0
0
30
60
Ncq
205
90
4
Ns 2
0
30
60
90
2
Ns 4
0
30
60
90
0
2
3
4
1
b for N 0; b for N 0
s
s
H
B
0
5
Figure 4.16 Meyerhof’s bearing capacity
factor Ncq for purely cohesive soil
B
Df
(a)
b
(b)
Figure 4.17 Schematic diagram of failure
zones for embedment and setback:
(a) Df /B 0; (b) b/B 0
206 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
1000
1000
b/B = 0
b/B = 0.5
b/B = 1
b/B = 2
= 45°
40°
40°
Nq
Nq
= 45°
100
100
35°
35°
30°
30°
10
0
10
20
30
10
40
0
10
20
(deg)
(deg)
(a)
(b)
30
40
Figure 4.18 Graham et al.’s theoretical values of Nq(Df/B 0)
1000
1000
b/B = 0
b/B = 0.5
= 45°
40°
Nq
Nq
40°
100
35°
35°
100
30°
30°
10
0
b/B = 1
b/B = 2
= 45°
10
20
30
40
10
0
10
20
(deg)
(deg)
(a)
(b)
Figure 4.19 Graham et al.’s theoretical values of Nq(Df/B 0.5)
30
40
4.6 Bearing Capacity of Foundations on Top of a Slope
207
1000
1000
= 45°
b/B = 0
b/B = 0.5
= 45°
b/B = 1
b/B = 2
40°
40°
Nq
Nq
35°
35°
100
100
30°
30°
10
0
10
20
30
10
40
0
10
20
(deg)
(deg)
(a)
(b)
30
40
Figure 4.20 Graham et al.’s theoretical values of Nq(Df/B 1)
Example 4.7
In Figure 4.14, for a shallow continuous foundation in a clay, the following data
are given: B 5 1.2 m; Df 5 1.2 m; b 5 0.8 m; H 5 6.2 m; b 5 30°; unit weight of
soil 5 17.5 kN>m3 ; f 5 0; and c 5 50 kN>m2 . Determine the gross allowable bearing
capacity with a factor of safety FS 5 4.
Solution
Since B , H, we will assume the stability number Ns 5 0. From Eq. (4.39),
qu 5 cNcq
We are given that
Df
B
5
1.2
51
1.2
and
b
0.8
5
5 0.67
B
1.2
208 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
For b 5 30°, Df>B 5 1 and b>B 5 0.67, Figure 4.16 gives Ncq 5 6.3. Hence,
qu 5 (50) (6.3) 5 315 kN>m2
and
qall 5
qu
315
5
5 78.8 kN , m2
FS
4
■
Example 4.8
Figure 4.21 shows a continuous foundation on a slope of a granular soil. Estimate the
ultimate bearing capacity.
1.5 m
1.5 m 1.5 m
6m
30º
16.8 kN/m3
40º
c 0
Figure 4.21 Foundation on a granular slope
Solution
For granular soil (cr 5 0), from Eq. (4.38),
qu 5 12 gBNgq
We are given that b>B 5 1.5>1.5 5 1, Df>B 5 1.5>1.5 5 1, fr 5 40°,
and b 5 30°.
From Figure 4.15, Ngq < 120. So,
qu 5 12 (16.8) (1.5) (120) 5 1512 kN , m2
■
Example 4.9
Solve Example 4.8 using the stress characteristics solution method.
Solution
1
qu 5 gBNgq
2
From Figure 4.20b, Nq ⬇ 110. Hence,
1
qu 5 (16.8) (1.5) (110) 5 1386 kN>m2
2
■
4.7 Seismic Bearing Capacity of a Foundation at the Edge of a Granular Soil Slope
4.7
209
Seismic Bearing Capacity of a Foundation at the Edge
of a Granular Soil Slope
Figure 4.22 shows a continuous surface foundation (B/L 0, Df/B 0) at the edge of a
granular slope. The foundation is subjected to a loading inclined at an angle to the vertical. Let the foundation be subjected to seismic loading with a horizontal coefficient of
acceleration, kh. Based on their analysis of method of slices, Huang and Kang (2008)
expressed the ultimate bearing capacity as
1
qu 5 gBNgFgiFgbFge
2
(4.41)
where
N bearing capacity factor (Table 3.3)
Fi load inclination factor
F slope inclination factor
Fe correction factor for the inertia force induced by seismic loading
The relationships for Fi, F , and Fe are as follow:
Fgi 5 c1 2 a
a° (0.1f921.21)
bd
fr°
(4.42)
Fgb 5 31 2 (1.062 2 0.014f9 )tanf r4 a 10° b
b°
(4.43)
and
Fe 1
[(2.57
0.043)e1.45 tan ]kh
(4.44)
In Eqs. (4.42) through (4.44), is in degrees.
qu
B
Sand
c = 0
Figure 4.22 Continuous
foundation at the edge of a
granular slope subjected to
seismic loading
210 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
Example 4.10
Consider a continuous surface foundation on a granular soil slope subjected to a seismic
loading, as shown in Figure 4.22. Given: B 1.5 m, 17.5 kN/m3, 35°, c 0, 30°, 10°, and kh 0.2. Calculate the ultimate bearing capacity, qu.
Solution
From Eq. (4.41),
1
qu 5 gBNgFgiFgbFge
2
For 35°, N 48.03 (Table 3.3). Thus,
Fgt 5 c1 2 a
a° (0.1f921.21)
10 3(0.1335)21.214
b
d
5
c1
2
a
bd
5 0.463
35
f9°
Fgb 5 31 2 (1.062 2 0.014f9 )tanf r4 a 10° b
b°
5 31 2 (1.062 2 0.014 3 35)tan 354 a 10 b 5 0.215
30
Fe 1
3(2.57
3(2.57
1
0.043)e1.45 tan 4 kh
0.043
35)e1.45 tan 304 (0.2) 0.508
So
1
qu 5 (17.5) (1.5) (48.03) (0.463) (0.215) (0.508) 5 31.9 kN>m2
2
4.8
■
Bearing Capacity of Foundations on a Slope
A theoretical solution for the ultimate bearing capacity of a shallow foundation located on
the face of a slope was developed by Meyerhof (1957). Figure 4.23 shows the nature of the
plastic zone developed under a rough continuous foundation of width B. In Figure 4.23,
abc is an elastic zone, acd is a radial shear zone, and ade is a mixed shear zone. Based on
this solution, the ultimate bearing capacity can be expressed as
B
qu
Df
a
b
e
90 d
90 c
Figure 4.23 Nature of plastic
zone under a rough continuous
foundation on the face of a slope
211
4.8 Bearing Capacity of Foundations on a Slope
qu 5 cNcqs (for purely cohesive soil, that is, f 5 0)
(4.45)
qu 5 12 gBNgqs (for granular soil, that is cr 5 0)
(4.46)
and
The variations of Ncqs and Ngqs with slope angle b are given in Figures 4.24 and 4.25.
8
Df /B 0
Df /B 1
7
6
Ns 0
Ncqs
5
4
0
3
1
2
2
3
4
1
5
5.53
0
0
20
40
60
(deg)
Figure 4.24 Variation of Ncqs with b.
(Note: Ns 5 gH>c)
80
600
500
Df /B 0
Df /B 1
400
300
45
Nqs
200
40
100
45
50
30
25
40
10
5
1
0
30
0
10
20
30
(deg)
40
50
Figure 4.25 Variation of Ngqs with b
212 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
4.9
Foundations on Rock
On some occasions, shallow foundations may have to be built on rocks, as shown in
Figure 4.26. For estimation of the ultimate bearing capacity of shallow foundations on
rock, we may use Terzaghi’s bearing capacity equations [Eqs. (3.3), (3.7) and (3.8)] with
the bearing capacity factors given here (Stagg and Zienkiewicz, 1968; Bowles, 1996):
f9
b
2
Nc 5 5 tan4 a45 1
Nq 5 tan6 a45 1
f9
b
2
N Nq 1
(4.47)
(4.48)
(4.49)
For rocks, the magnitude of the cohesion intercept, c, can be expressed as
quc 5 2c9tana45 1
f9
b
2
(4.50)
where
quc unconfined compression strength of rock
angle of friction
The unconfined compression strength and the friction angle of rocks can vary widely.
Table 4.2 gives a general range of quc for various types of rocks. It is important to keep in mind
that the magnitude of quc and (hence c) reported from laboratory tests are for intact rock
specimens. It does not account for the effect of discontinuities. To account for discontinuities,
Bowles (1996) suggested that the ultimate bearing capacity qu should be modified as
qu(modified) qu(RQD)2
where RQD rock quality designation (see Chapter 2).
γ
Soil
Df
B
Rock
c
Figure 4.26 Foundation on rock
Table 4.2 Range of the Unconfined Compression Strength
of Various Types of Rocks
quc
Rock type
MN/m2
kip/m2
(deg)
Granite
Limestone
Sandstone
Shale
65–250
30–150
25–130
5–40
9.5–36
4–22
3.5–19
0.75–6
45–55
35–45
30–45
15–30
(4.51)
4.10 Uplift Capacity of Foundations
213
In any case, the upper limit of the allowable bearing capacity should not exceed fc
(28-day compressive strength of concrete).
Example 4.11
Refer to Figure 4.26. A square column foundation is to be constructed over siltstone.
Given:
B 2.5 m 2.5 m
Df 2 m
17 kN/m3
c 32 MN/m2
31°
25 kN/m3
RDQ 50%
Foundation: B
Soil:
Siltstone:
Estimate the allowable load-bearing capacity. Use FS 4. Also, for concrete, use fc 30 MN/m2.
Solution
From Eq. (3.7),
qu 1.3cNc qNq 0.4BN
Nc 5 5 tan4 a45 1
Nq 5 tan6 a45 1
f9
31
b 5 5 tan4 a45 1 b 5 48.8
2
2
f9
31
b 5 tan6 a45 1 b 5 30.5
2
2
N Nq 1 30.5 1 31.5
Hence,
qu (1.3)(32
2030.08
2031.9
103 kN/m2)(48.8) (17
103 1.037
2)(30.5) (0.4)(25)(2.5)(31.5)
103 0.788
103
103 kN/m2 ⬇ 2032 MN/m2
qu(modified) qu(RQD)2 (2032)(0.5)2 508 MN/m2
q all 5
508
5 127 MN>m2
4
Since 127 MN/m2 is greater than fc, use qall 30 MN/m2.
4.10
■
Uplift Capacity of Foundations
Foundations may be subjected to uplift forces under special circumstances. During the
design process for those foundations, it is desirable to provide a sufficient factor of safety
against failure by uplift. This section will provide the relationships for the uplift capacity
of foundations in granular and cohesive soils.
214 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
Foundations in Granular Soil (c9 5 0)
Figure 4.27 shows a shallow continuous foundation that is being subjected to an uplift
force. At ultimate load, Qu , the failure surface in soil will be as shown in the figure. The
ultimate load can be expressed in the form of a nondimensional breakout factor, Fq. Or
Fq 5
Qu
AgDf
(4.52)
where A 5 area of the foundation.
The breakout factor is a function of the soil friction angle fr and Df>B. For
a given soil friction angle, Fq increases with Df>B to a maximum at Df>B 5 (Df>B) cr
and remains constant thereafter. For foundations subjected to uplift, Df>B # (Df>B) cr is
considered a shallow foundation condition. When a foundation has an embedment ratio of
Df>B . (Df>B) cr , it is referred to as a deep foundation. Meyerhof and Adams (1968)
provided relationships to estimate the ultimate uplifting load Qu for shallow
3that is, Df>B # (Df>B) cr4 , circular, and rectangular foundations. Using these relationships and Eq. (4.52), Das and Seeley (1975) expressed the breakout factor Fq in the following form
Fq 5 1 1 2B1 1 m¢
Df
B
≤Ra
Df
B
bKu tan fr
(4.53)
(for shallow circular and square foundations)
Fq 5 1 1 b B 1 1 2m¢
Df
B
≤R¢
Df
B
≤ 1 1r ¢ ≤Ku tan fr
L
B
(for shallow rectangular foundations)
Qu
Pp
Pp
Sand
Unit weight γ
Friction angle Df
B
Figure 4.27 Shallow continuous foundation subjected to uplift
(4.54)
4.10 Uplift Capacity of Foundations
215
where
m 5 a coefficient which is a function of fr
Ku 5 nominal uplift coefficient
The variations of Ku, m, and (Df>B) cr for square and circular foundations are given
in Table 4.3 (Meyerhof and Adams, 1968).
For rectangular foundations, Das and Jones (1982) recommended that
¢
Df
B
≤
5 ¢
cr-rectangular
Df
B
≤
B0.133¢
cr-square
Df
L
≤ 1 0.867R # 1.4¢ ≤
B
B cr-square
(4.55)
Using the values of Ku, m, and (Df>B) cr in Eq. (4.53), the variations of Fq for square
and circular foundations have been calculated and are shown in Figure 4.28. A step-bystep procedure to estimate the uplift capacity of foundations in granular soil follows.
Determine, Df, B, L, and fr.
CalculateDf>B.
Using Table 4.3 and Eq. (4.55), calculate (Df >B) cr.
If Df>B is less than or equal to (Df>B) cr, it is a shallow foundation.
If Df>B . (Df>B) cr, it is a deep foundation.
For shallow foundations, use Df>B calculated in Step 2 in Eq. (4.53) or
(4.54) to estimate Fq. Thus, Qu 5 Fq AgDf.
Step 7. For deep foundations, substitute (Df>B) cr for Df>B in Eq. (4.53) or (4.54)
to obtain Fq, from which the ultimate load Qu may be obtained.
Step 1.
Step 2.
Step 3.
Step 4.
Step 5.
Step 6.
100
45
40
Fq
35
10
30
1
1
2
3
4
5
6
Df /B
7
8
9
10
Figure 4.28 Variation of Fq with
Df>B and fr
216 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
Table 4.3 Variation of Ku, m, and (Df>B) cr
Soil friction
angle, f9 (deg)
Ku
m
(Df >B)cr for square
and circular foundations
20
25
30
35
40
45
0.856
0.888
0.920
0.936
0.960
0.960
0.05
0.10
0.15
0.25
0.35
0.50
2.5
3
4
5
7
9
Foundations in Cohesive Soil (f9 5 0)
The ultimate uplift capacity, Qu, of a foundation in a purely cohesive soil can be expressed as
Qu 5 A(gDf 1 cuFc )
(4.56)
where
A 5 area of the foundation
cu 5 undrained shear strength of soil
Fc 5 breakout factor
As in the case of foundations in granular soil, the breakout factor Fc increases with embedment ratio and reaches a maximum value of Fc 5 F c* at Df>B 5 (Df>B) cr and remains
constant thereafter.
Das (1978) also reported some model test results with square and rectangular foundations. Based on these test results, it was proposed that
¢
Df
B
≤
5 0.107cu 1 2.5 # 7
(4.57)
cr-square
where
Df
critical embedment ratio of square (or circular) foundations
¢ ≤
B cr-square
cu 5 undrained cohesion, in kN>m2
It was also observed by Das (1980) that
¢
Df
B
≤
5 ¢
cr-rectangular
Df
B
≤
B0.73 1 0.27¢
cr-square
Df
L
(4.58)
≤ R # 1.55¢ ≤
B
B cr-square
where
Df
¢ ≤
critical embedment ratio of rectangular foundations
B cr-rectangular
L 5 length of foundation
217
4.10 Uplift Capacity of Foundations
Based on these findings, Das (1980) proposed an empirical procedure to obtain the
breakout factors for shallow and deep foundations. According to this procedure, ar and br
are two nondimensional factors defined as
Df
ar 5
¢
B
Df
B
(4.59)
≤
cr
and
br 5
Fc
(4.60)
F c*
For a given foundation, the critical embedment ratio can be calculated using Eqs. (4.57)
and (4.58). The magnitude of F c* can be given by the following empirical relationship
*
F c-rectangular
5 7.56 1 1.44¢
B
≤
L
(4.61)
*
5 breakout factor for deep rectangular foundations
where F c-rectangular
Figure 4.29 shows the experimentally derived plots (upper limit, lower limit, and
average of br and ar. Following is a step-by-step procedure to estimate the ultimate uplift
capacity.
Step 1.
Step 2.
Step 3.
Step 4.
Determine the representative value of the undrained cohesion, cu.
Determine the critical embedment ratio using Eqs. (4.57) and (4.58).
Determine the Df>B ratio for the foundation.
If Df>B . (Df>B) cr, as determined in Step 2, it is a deep foundation.
However, if Df>B # (Df>B) cr, it is a shallow foundation.
1.2
1.0
0.8
p
l
er
Up
0.6
im
it
ge it
era lim
v
A er
w
Lo
0.4
0.2
0
0
0.2
0.4
Figure 4.29 Plot of br versus ar
0.6
0.8
1.0
218 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
Step 5. For Df>B . (Df>B) cr
Fc 5 F c* 5 7.56 1 1.44¢
B
≤
L
Thus,
Qu 5 Ab B 7.56 1 1.44¢
B
≤ Rcu 1 gDf r
L
(4.62)
where A area of the foundation.
Step 6. For Df>B # (Df>B) cr
Qu 5 A(brF c* cu 1 gDf ) 5 Abbr B7.56 1 1.44¢
B
≤ R cu 1 gDf r (4.63)
L
The value of br can be obtained from the average curve of Figure 4.29. The procedure
outlined above gives fairly good results for estimating the net ultimate uplift capacity
of foundations and agrees reasonably well with the theoretical solution of Merifield
et al. (2003).
Example 4.12
Consider a circular foundation in sand. Given for the foundation: diameter, B 5 1.5 m and
depth of embedment, Df 5 1.5 m. Given for the sand: unit weight, g 5 17.4 kN>m3 , and
friction angle, fr 5 35°. Calculate the ultimate bearing capacity.
Solution
Df>B 5 1.5>1.5 5 1 and fr 5 35°. For circular foundation, (Df>B) cr 5 5. Hence, it
is a shallow foundation. From Eq. (4.53)
Fq 5 1 1 2B1 1 m¢
Df
B
≤R¢
Df
B
≤ Ku tan fr
For fr 5 35°, m 0.25, and Ku 0.936 (Table 4.3). So
Fq 5 1 1 231 1 (0.25) (1)4 (1) (0.936) (tan35) 5 2.638
So
p
Qu 5 FqgADf 5 (2.638) (17.4) B ¢ ≤ (1.5) 2 R (1.5) 5 121.7 kN
4
■
Problems
219
Example 4.13
A rectangular foundation in a saturated clay measures 1.5 m
3 m. Given:
Df 5 1.8 m, cu 5 52 kN>m2 , and g 5 18.9 kN>m3 . Estimate the ultimate uplift capacity.
Solution
From Eq. (4.57)
¢
Df
B
≤
5 0.107cu 1 2.5 5 (0.107) (52) 1 2.5 5 8.06
cr-square
So use (Df>B) cr-square 5 7 . Again from Eq. (4.58),
Df
Df
L
5 ¢ ≤
B0.73 1 0.27¢ ≤ R
¢ ≤
B cr-rectangular
B cr-square
B
5 7B0.73 1 0.27¢
Check:
1.55¢
Df
B
≤
3
≤ R 5 8.89
1.5
5 (1.55) (7) 5 10.85
cr-square
So use (Df>B) cr-rectangular 5 8.89 . The actual embedment ratio is Df>B 5 1.8>1.5 5 1.2.
Hence, this is a shallow foundation.
Df
ar 5
¢
B
Df
B
5
≤
1.2
5 0.135
8.89
cr
Referring to the average curve of Figure 4.29, for ar 5 0.135, the magnitude of
br 5 0.2. From Eq. (4.63),
Qu 5 Abbr B7.56 1 1.44¢
B
≤ R cu 1 gDf r
L
5 (1.5) (3) b (0.2) B7.56 1 1.44¢
1.5
≤ R (52) 1 (18.9) (1.8) r 5 540.6 kN
3
■
Problems
4.1
Refer to Figure 4.2 and consider a rectangular foundation. Given: B 0.91 m, L 1.83 m, Df 0.91 m, H 0.61 m, 40°, c 0, and 18.08 kN/m3. Using a
factor of safety of 4, determine the gross allowable load the foundation can carry.
Use Eq. (4.3).
220 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
Repeat Problem 4.1 with the following data: B 1.5 m, L 1.5 m, Df 1 m, H 0.6 m, 35°, c 0, and 15 kN/m3. Use FS 3.
Refer to Figure 4.2. Given: B L 1.75 m, Df 1 m, H 1.75 m, 17 kN/m3,
c 0, and 30°. Using Eq. (4.6) and FS 4, determine the gross allowable
load the foundation can carry.
Refer to Figure 4.2. A square foundation measuring 1.22 m 1.22 m is supported by a
saturated clay layer of limited depth underlain by a rock layer. Given that Df 0.91 m,
H 0.61 m, cu 115 kN/m2, and 18.87 kN/m3, estimate the ultimate bearing
capacity of the foundation.
Refer to Figure 4.8. For a strip foundation in two-layered clay, given:
• 1 18.08 kN/m3, c1 57.5 kN/m2, 1 0
• 2 17.29 kN/m3, c2 28.75 kN/m2, 2 0
• B 0.91 m, Df 0.61 m, H 0.61 m
Find the gross allowable bearing capacity. Use a factor of safety of 3.
Refer to Figure 4.8. For a strip foundation in two-layered clay, given:
• B 0.92 m, L 1.22 m, Df 0.92 m, H 0.76 m
• 1 17 kN/m3, 1 0, c1 72 kN/m2
• 2 17 kN/m3, 2 0, c2 43 kN/m2
Determine the gross ultimate bearing capacity.
Refer to Figure 4.8. For a square foundation on layered sand, given:
• B 1.5 m, Df 1.5 m, H 1 m
• 1 18 kN/m3, 1 40°, c1 0
• 2 16.7 kN/m3, 2 32°, c2 0
Determine the net allowable load that the foundations can support. Use FS 4.
Refer to Figure 4.11. For a rectangular foundation on layered sand, given:
• B 1.22 m, L 1.83 m, H 0.61 m, Df 0.91 m
• 1 15.41 kN/m3, 1 30°, c1 0
• 2 16.98 kN/m3, 2 38°, c2 0
Using a factor of safety of 4, determine the gross allowable load the foundation
can carry.
Two continuous shallow foundations are constructed alongside each other in a granular soil. Given, for the foundation: B 1.2 m, Df 1 m, and center-to-center spacing 2 m. The soil friction angle, 35°. Estimate the net allowable bearing
capacity of the foundations. Use a factor of safety of FS 4 and a unit weight of soil
g 5 16.8 kN>m3.
A continuous foundation with a width of 1 m is located on a slope made of clay soil.
Refer to Figure 4.14 and let Df 1 m, H 4 m, b 2 m, g 5 16.8 kN>m3, c 68 kN/m2 , 0, and 60°.
a. Determine the allowable bearing capacity of the foundation. Let FS 3.
b. Plot a graph of the ultimate bearing capacity qu if b is changed from 0 to 6 m.
A continuous foundation is to be constructed near a slope made of granular soil
(see Figure 4.14). If B 1.22 m, b 1.83 m, H 4.57 m, Df 1.22 m, b 30°,
fr 5 40°, and g 5 17.29 kN>m3, estimate the ultimate bearing capacity of the
foundation. Use Meyerhof’s solution.
A square foundation in a sand deposit measures 1.22 m
1.22 m in plan. Given:
Df 1.52 m, soil friction angle 5 35°, and unit weight of soil 5 17.6 kN>m3. Estimate the ultimate uplift capacity of the foundation.
References
221
4.13 A foundation measuring 1.2 m
2.4 m in plan is constructed in a saturated clay.
Given: depth of embedment of the foundation 2 m, unit weight of soil 18 kN>m3 , and undrained cohesion of clay 74 kN>m2 . Estimate the ultimate uplift
capacity of the foundation.
References
BOWLES, J. E. (1996). Foundation Analysis and Design, 5th. edition, McGraw-Hill, New York.
BUISMAN, A. S. K. (1940). Grondmechanica, Waltman, Delft, the Netherlands.
CERATO, A. B., and LUTENEGGER, A. J. (2006). “Bearing Capacity of Square and Circular Footings on a Finite Layer of Granular Soil Underlain by a Rigid Base,” Journal of Geotechnical
and Geoenvironmental Engineering, American Society of Civil Engineers, Vol. 132, No. 11,
pp. 1496–1501.
DAS, B. M. (1978). “Model Tests for Uplift Capacity of Foundations in Clay,” Soils and Foundations, Vol. 18, No. 2, pp. 17–24.
DAS, B. M. (1980). “A Procedure for Estimation of Ultimate Uplift Capacity of Foundations in
Clay,” Soils and Foundations, Vol. 20, No. 1, pp. 77–82.
DAS, B. M., and JONES, A. D. (1982). “Uplift Capacity of Rectangular Foundations in Sand,”
Transportation Research Record 884, National Research Council, Washington, D.C.,
pp. 54–58.
DAS, B. M., and SEELEY, G. R. (1975). “Breakout Resistance of Horizontal Anchors,” Journal of
Geotechnical Engineering Division, ASCE, Vol. 101, No. 9, pp. 999–1003.
GRAHAM, J., ANDREWS, M., and SHIELDS, D. H. (1988). “Stress Characteristics for Shallow
Footing on Cohesionless Slopes,” Canadian Geotechnical Journal, Vol. 25, No. 2,
pp. 238–249.
HUANG, C. C., and KANG, W. W. (2008). “Seismic Bearing Capacity of a Rigid Footing Adjacent
to a Cohesionless Slope,” Soils and Foundations, Vol. 48, No. 5, pp. 641–651.
LUNDGREN, H., and MORTENSEN, K. (1953). “Determination by the Theory of Plasticity
on the Bearing Capacity of Continuous Footings on Sand,” Proceedings, Third International Conference on Soil Mechanics and Foundation Engineering, Zurich, Vol. 1,
pp. 409 – 412.
MANDEL, J., and SALENCON, J. (1972). “Force portante d’un sol sur une assise rigide (étude
théorique),” Geotechnique, Vol. 22, No. 1, pp. 79–93.
MERIFIELD, R. S., LYAMIN, A., and SLOAN, S. W. (2003). “Three Dimensional Lower Bound
Solutions for the Stability of Plate Anchors in Clay,” Journal of Geotechnical and Geoenvironmental Engineering, ASCE, Vo. 129, No. 3, pp. 243–253.
MEYERHOF, G. G. (1957). “The Ultimate Bearing Capacity of Foundations on Slopes,” Proceedings, Fourth International Conference on Soil Mechanics and Foundation Engineering,
London, Vol. 1, pp. 384 –387.
MEYERHOF, G. G. (1974). “Ultimate Bearing Capacity of Footings on Sand Layer Overlying Clay,”
Canadian Geotechnical Journal, Vol. 11, No. 2, pp. 224– 229.
MEYERHOF, G. G., and ADAMS, J. I. (1968). “The Ultimate Uplift Capacity of Foundations,”
Canadian Geotechnical Journal, Vol. 5, No. 4, pp. 225–244.
MEYERHOF, G. G., and CHAPLIN, T. K. (1953). “The Compression and Bearing Capacity of
Cohesive Soils,” British Journal of Applied Physics, Vol. 4, pp. 20–26.
MEYERHOF, G. G., and HANNA, A. M. (1978). “Ultimate Bearing Capacity of Foundations on
Layered Soil under Inclined Load,” Canadian Geotechnical Journal, Vol. 15, No. 4, pp.
565– 572.
222 Chapter 4: Ultimate Bearing Capacity of Shallow Foundations: Special Cases
PRANDTL, L. (1921). “Über die Eindringungsfestigkeit (Härte) plastischer Baustoffe und die Festigkeit von Schneiden,” Zeitschrift für angewandte Mathematik und Mechanik, Vol. 1, No. 1,
pp. 15– 20.
REISSNER, H. (1924). “Zum Erddruckproblem,” Proceedings, First International Congress of
Applied Mechanics, Delft, the Netherlands, pp. 295–311.
STAGG, K. G., and ZIENKIEWICZ, O. C. (1968). Rock Mechanics in Engineering Practice, John
Wiley & Sons, New York.
5
5.1
Shallow Foundations:
Allowable Bearing Capacity and Settlement
Introduction
It was mentioned in Chapter 3 that, in many cases, the allowable settlement of a shallow
foundation may control the allowable bearing capacity. The allowable settlement itself
may be controlled by local building codes. Thus, the allowable bearing capacity will be the
smaller of the following two conditions:
qu
FS
qall 5 d or
qallowable settlement
The settlement of a foundation can be divided into two major categories: (a) elastic, or
immediate, settlement and (b) consolidation settlement. Immediate, or elastic, settlement
of a foundation takes place during or immediately after the construction of the structure.
Consolidation settlement occurs over time. Pore water is extruded from the void spaces of
saturated clayey soils submerged in water. The total settlement of a foundation is the sum
of the elastic settlement and the consolidation settlement.
Consolidation settlement comprises two phases: primary and secondary. The
fundamentals of primary consolidation settlement were explained in detail in Chapter 1.
Secondary consolidation settlement occurs after the completion of primary consolidation
caused by slippage and reorientation of soil particles under a sustained load. Primary
consolidation settlement is more significant than secondary settlement in inorganic clays
and silty soils. However, in organic soils, secondary consolidation settlement is more
significant.
For the calculation of foundation settlement (both elastic and consolidation), it is
required that we estimate the vertical stress increase in the soil mass due to the net load
applied on the foundation. Hence, this chapter is divided into the following three parts:
1. Procedure for calculation of vertical stress increase
2. Elastic settlement calculation
3. Consolidation settlement calculation
223
224 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Vertical Stress Increase in a Soil Mass Caused
by Foundation Load
5.2
Stress Due to a Concentrated Load
In 1885, Boussinesq developed the mathematical relationships for determining the normal and
shear stresses at any point inside homogeneous, elastic, and isotropic mediums due to a concentrated point load located at the surface, as shown in Figure 5.1. According to his analysis,
the vertical stress increase at point A caused by a point load of magnitude P is given by
Ds 5
3P
r 2 5>2
2pz B1 1 ¢ ≤ R
z
(5.1)
2
where
r 5 "x2 1 y2
x, y, z 5 coordinates of the point A
Note that Eq. (5.1) is not a function of Poisson’s ratio of the soil.
5.3
Stress Due to a Circularly Loaded Area
The Boussinesq equation (5.1) can also be used to determine the vertical stress below the
center of a flexible circularly loaded area, as shown in Figure 5.2. Let the radius of the
loaded area be B>2, and let qo be the uniformly distributed load per unit area. To determine
P
x
r
y
z
A
(x,y,z)
Figure 5.1 Vertical stress at a point A
caused by a point load on the surface
5.3 Stress Due to a Circularly Loaded Area
225
qo
B/2
r
r
dr
d
qo
z
z
A
A
Figure 5.2 Increase in pressure under a uniformly
loaded flexible circular area
the stress increase at a point A, located at a depth z below the center of the circular area,
consider an elemental area on the circle. The load on this elemental area may be taken to
be a point load and expressed as qor du dr. The stress increase at A caused by this load can
be determined from Eq. (5.1) as
3(qor du dr)
ds 5
(5.2)
r 2 5>2
2pz B1 1 ¢ ≤ R
z
2
The total increase in stress caused by the entire loaded area may be obtained by integrating Eq. (5.2), or
u52p
Ds 5 3ds 5 3
r5B>2
3
u50
5 qo d1 2
r50
3(qor du dr)
r 2 5>2
2pz2 B1 1 ¢ ≤ R
z
1
B 2 3>2
B1 1 ¢ ≤ R
2z
t
(5.3)
Similar integrations could be performed to obtain the vertical stress increase at
Ar, located a distance r from the center of the loaded area at a depth z (Ahlvin and
Ulery, 1962). Table 5.1 gives the variation of Ds>qo with r> (B>2) and z> (B>2) [for
0 # r> (B>2) # 1]. Note that the variation of Ds>qo with depth at r> (B>2) 5 0 can be
obtained from Eq. (5.3).
226 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Table 5.1 Variation of Ds>qo for a Uniformly Loaded Flexible Circular Area
r , (B , 2)
5.4
z , (B , 2)
0
0.2
0.4
0.6
0.8
1.0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.5
2.0
2.5
3.0
4.0
1.000
0.999
0.992
0.976
0.949
0.911
0.864
0.811
0.756
0.701
0.646
0.546
0.424
0.286
0.200
0.146
0.087
1.000
0.999
0.991
0.973
0.943
0.902
0.852
0.798
0.743
0.688
0.633
0.535
0.416
0.286
0.197
0.145
0.086
1.000
0.998
0.987
0.963
0.920
0.869
0.814
0.756
0.699
0.644
0.591
0.501
0.392
0.268
0.191
0.141
0.085
1.000
0.996
0.970
0.922
0.860
0.796
0.732
0.674
0.619
0.570
0.525
0.447
0.355
0.248
0.180
0.135
0.082
1.000
0.976
0.890
0.793
0.712
0.646
0.591
0.545
0.504
0.467
0.434
0.377
0.308
0.224
0.167
0.127
0.080
1.000
0.484
0.468
0.451
0.435
0.417
0.400
0.367
0.366
0.348
0.332
0.300
0.256
0.196
0.151
0.118
0.075
Stress below a Rectangular Area
The integration technique of Boussinesq’s equation also allows the vertical stress at any
point A below the corner of a flexible rectangular loaded area to be evaluated. (See
Figure 5.3.) To do so, consider an elementary area dA 5 dx dy on the flexible loaded
area. If the load per unit area is qo , the total load on the elemental area is
dP 5 qo dx dy
(5.4)
This elemental load, dP, may be treated as a point load. The increase in vertical stress at
point A caused by dP may be evaluated by using Eq. (5.1). Note, however, the need to substitute dP 5 qo dx dy for P and x2 1 y2 for r2 in that equation. Thus,
The stress increase at A caused by dP 5
3qo (dx dy)z3
2p(x2 1 y2 1 z2 ) 5>2
The total stress increase Ds caused by the entire loaded area at point A may now be obtained by integrating the preceding equation:
L
Ds 5 3
B
3qo (dx dy)z3
3
2
y50 x50 2p(x
1 y2 1 z2 ) 5>2
5 qoI
(5.5)
227
5.4 Stress below a Rectangular Area
x
qo
B
dx
dy
y
L
z
Figure 5.3 Determination of
stress below the corner of a flexible
rectangular loaded area
A
Here,
I 5 influence factor 5
2mn"m2 1 n2 1 1 # m2 1 n2 1 2
1
¢ 2
4p m 1 n2 1 m2n2 1 1 m2 1 n2 1 1
1 tan21
2mn"m2 1 n2 1 1
≤
m2 1 n2 1 1 2 m2n2
(5.6)
where
m5
B
z
(5.7)
n5
L
z
(5.8)
and
The variations of the influence values with m and n are given in Table 5.2.
The stress increase at any point below a rectangular loaded area can also be found by
using Eq. (5.5) in conjunction with Figure 5.4. To determine the stress at a depth z below point
O, divide the loaded area into four rectangles, with O the corner common to each. Then use
Eq. (5.5) to calculate the increase in stress at a depth z below O caused by each rectangular
area. The total stress increase caused by the entire loaded area may now be expressed as
Ds 5 qo (I1 1 I2 1 I3 1 I4 )
(5.9)
where I1 , I2 , I3 , and I4 5 the influence values of rectangles 1, 2, 3, and 4, respectively.
In most cases, the vertical stress below the center of a rectangular area is of importance. This can be given by the relationship
Ds 5 qoIc
(5.10)
228
0.1
0.00470
0.00917
0.01323
0.01678
0.01978
0.02223
0.02420
0.02576
0.02698
0.02794
0.02926
0.03007
0.03058
0.03090
0.03111
0.03138
0.03150
0.03158
0.03160
0.03161
0.03162
0.03162
0.03162
m
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
1.6
1.8
2.0
2.5
3.0
4.0
5.0
6.0
8.0
10.0
`
0.06201
0.06202
0.06202
0.06202
0.06155
0.06178
0.06194
0.06199
0.05733
0.05894
0.05994
0.06058
0.06100
0.04348
0.04735
0.05042
0.05283
0.05471
0.00917
0.01790
0.02585
0.03280
0.03866
0.2
0.09017
0.09018
0.09019
0.09019
0.08948
0.08982
0.09007
0.09014
0.08323
0.08561
0.08709
0.08804
0.08867
0.06294
0.06858
0.07308
0.07661
0.07938
0.01323
0.02585
0.03735
0.04742
0.05593
0.3
0.11541
0.11543
0.11544
0.11544
0.11450
0.11495
0.11527
0.11537
0.10631
0.10941
0.11135
0.11260
0.11342
0.08009
0.08734
0.09314
0.09770
0.10129
0.01678
0.03280
0.04742
0.06024
0.07111
0.4
Table 5.2 Variation of Influence Value I [Eq. (5.6)]a
0.13741
0.13744
0.13745
0.13745
0.13628
0.13684
0.13724
0.13737
0.12626
0.13003
0.13241
0.13395
0.13496
0.09473
0.10340
0.11035
0.11584
0.12018
0.01978
0.03866
0.05593
0.07111
0.08403
0.5
0.15617
0.15621
0.15622
0.15623
0.15483
0.15550
0.15598
0.15612
0.14309
0.14749
0.15028
0.15207
0.15326
0.10688
0.11679
0.12474
0.13105
0.13605
0.02223
0.04348
0.06294
0.08009
0.09473
0.6
n
0.17191
0.17195
0.17196
0.17197
0.17036
0.17113
0.17168
0.17185
0.15703
0.16199
0.16515
0.16720
0.16856
0.11679
0.12772
0.13653
0.14356
0.14914
0.02420
0.04735
0.06858
0.08734
0.10340
0.7
0.18496
0.18500
0.18502
0.18502
0.18321
0.18407
0.18469
0.18488
0.16843
0.17389
0.17739
0.17967
0.18119
0.12474
0.13653
0.14607
0.15371
0.15978
0.02576
0.05042
0.07308
0.09314
0.11035
0.8
0.19569
0.19574
0.19576
0.19577
0.19375
0.19470
0.19540
0.19561
0.17766
0.18357
0.18737
0.18986
0.19152
0.13105
0.14356
0.15371
0.16185
0.16835
0.02698
0.05283
0.07661
0.09770
0.11584
0.9
0.20449
0.20455
0.20457
0.20458
0.20236
0.20341
0.20417
0.20440
0.18508
0.19139
0.19546
0.19814
0.19994
0.13605
0.14914
0.15978
0.16835
0.17522
0.02794
0.05471
0.07938
0.10129
0.12018
1.0
0.21760
0.21767
0.21769
0.21770
0.21512
0.21633
0.21722
0.21749
0.19584
0.20278
0.20731
0.21032
0.21235
0.14309
0.15703
0.16843
0.17766
0.18508
0.02926
0.05733
0.08323
0.10631
0.12626
1.2
0.22644
0.22652
0.22654
0.22656
0.22364
0.22499
0.22600
0.22632
0.20278
0.21020
0.21510
0.21836
0.22058
0.14749
0.16199
0.17389
0.18357
0.19139
0.03007
0.05894
0.08561
0.10941
0.13003
1.4
229
a
0.03058
0.05994
0.08709
0.11135
0.13241
0.15028
0.16515
0.17739
0.18737
0.19546
0.20731
0.21510
0.22025
0.22372
0.22610
0.22940
0.23088
0.23200
0.23236
0.23249
0.23258
0.23261
0.23263
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
1.6
1.8
2.0
2.5
3.0
4.0
5.0
6.0
8.0
10.0
`
After Newmark, 1935.
1.6
m
1.8
0.23671
0.23681
0.23684
0.23686
0.23334
0.23495
0.23617
0.23656
0.21032
0.21836
0.22372
0.22736
0.22986
0.15207
0.16720
0.17967
0.18986
0.19814
0.03090
0.06058
0.08804
0.11260
0.13395
Table 5.2 (Continued)
0.23970
0.23981
0.23985
0.23987
0.23614
0.23782
0.23912
0.23954
0.21235
0.22058
0.22610
0.22986
0.23247
0.15326
0.16856
0.18119
0.19152
0.19994
0.03111
0.06100
0.08867
0.11342
0.13496
2.0
0.24412
0.24425
0.24429
0.24432
0.24010
0.24196
0.24344
0.24392
0.21512
0.22364
0.22940
0.23334
0.23614
0.15483
0.17036
0.18321
0.19375
0.20236
0.03138
0.06155
0.08948
0.11450
0.13628
2.5
0.24630
0.24646
0.24650
0.24654
0.24196
0.24394
0.24554
0.24608
0.21633
0.22499
0.23088
0.23495
0.23782
0.15550
0.17113
0.18407
0.19470
0.20341
0.03150
0.06178
0.08982
0.11495
0.13684
3.0
0.24817
0.24836
0.24842
0.24846
0.24344
0.24554
0.24729
0.24791
0.21722
0.22600
0.23200
0.23617
0.23912
0.15598
0.17168
0.18469
0.19540
0.20417
0.03158
0.06194
0.09007
0.11527
0.13724
4.0
n
0.24885
0.24907
0.24914
0.24919
0.24392
0.24608
0.24791
0.24857
0.21749
0.22632
0.23236
0.23656
0.23954
0.15612
0.17185
0.18488
0.19561
0.20440
0.03160
0.06199
0.09014
0.11537
0.13737
5.0
0.24916
0.24939
0.24946
0.24952
0.24412
0.24630
0.24817
0.24885
0.21760
0.22644
0.23249
0.23671
0.23970
0.15617
0.17191
0.18496
0.19569
0.20449
0.03161
0.06201
0.09017
0.11541
0.13741
6.0
0.24939
0.24964
0.24973
0.24980
0.24425
0.24646
0.24836
0.24907
0.21767
0.22652
0.23258
0.23681
0.23981
0.15621
0.17195
0.18500
0.19574
0.20455
0.03162
0.06202
0.09018
0.11543
0.13744
8.0
0.24946
0.24973
0.24981
0.24989
0.24429
0.24650
0.24842
0.24914
0.21769
0.22654
0.23261
0.23684
0.23985
0.15622
0.17196
0.18502
0.19576
0.20457
0.03162
0.06202
0.09019
0.11544
0.13745
10.0
0.24952
0.24980
0.24989
0.25000
0.24432
0.24654
0.24846
0.24919
0.21770
0.22656
0.23263
0.23686
0.23987
0.15623
0.17197
0.18502
0.19577
0.20458
0.03162
0.06202
0.09019
0.11544
0.13745
ⴥ
230 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
B(1)
1
3
O
B(2)
2
4
L(1)
L(2)
Figure 5.4 Stress below any point of
a loaded flexible rectangular area
where
Ic 5
m1n1
1 1 m21 1 2n21
2
B
p "1 1 m2 1 n2 (1 1 n21 ) (m21 1 n21 )
1
1
1 sin21
m1 5
n1 5
L
B
"m21
m1
1 n21"1 1 n21
R
(5.11)
(5.12)
z
(5.13)
B
¢ ≤
2
The variation of Ic with m1 and n1 is given in Table 5.3.
Table 5.3 Variation of Ic with m1 and n1
m1
ni
1
2
3
4
5
6
7
8
9
10
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
1.80
2.00
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
0.994
0.960
0.892
0.800
0.701
0.606
0.522
0.449
0.388
0.336
0.179
0.108
0.072
0.051
0.038
0.029
0.023
0.019
0.997
0.976
0.932
0.870
0.800
0.727
0.658
0.593
0.534
0.481
0.293
0.190
0.131
0.095
0.072
0.056
0.045
0.037
0.997
0.977
0.936
0.878
0.814
0.748
0.685
0.627
0.573
0.525
0.348
0.241
0.174
0.130
0.100
0.079
0.064
0.053
0.997
0.977
0.936
0.880
0.817
0.753
0.692
0.636
0.585
0.540
0.373
0.269
0.202
0.155
0.122
0.098
0.081
0.067
0.997
0.977
0.937
0.881
0.818
0.754
0.694
0.639
0.590
0.545
0.384
0.285
0.219
0.172
0.139
0.113
0.094
0.079
0.997
0.977
0.937
0.881
0.818
0.755
0.695
0.640
0.591
0.547
0.389
0.293
0.229
0.184
0.150
0.125
0.105
0.089
0.997
0.977
0.937
0.881
0.818
0.755
0.695
0.641
0.592
0.548
0.392
0.298
0.236
0.192
0.158
0.133
0.113
0.097
0.997
0.977
0.937
0.881
0.818
0.755
0.696
0.641
0.592
0.549
0.393
0.301
0.240
0.197
0.164
0.139
0.119
0.103
0.997
0.977
0.937
0.881
0.818
0.755
0.696
0.641
0.593
0.549
0.394
0.302
0.242
0.200
0.168
0.144
0.124
0.108
0.997
0.977
0.937
0.881
0.818
0.755
0.696
0.642
0.593
0.549
0.395
0.303
0.244
0.202
0.171
0.147
0.128
0.112
5.4 Stress below a Rectangular Area
231
qo
Foundation B L
2 vertical to
1 horizontal
B
2 vertical to
1 horizontal
z
Bz
Figure 5.5 2:1 method of finding stress increase under a foundation
Foundation engineers often use an approximate method to determine the increase
in stress with depth caused by the construction of a foundation. The method is referred to
as the 2:1 method. (See Figure 5.5.) According to this method, the increase in stress at
depth z is
Ds 5
qo 3 B 3 L
(B 1 z) (L 1 z)
(5.14)
`
Note that Eq. (5.14) is based on the assumption that the stress from the foundation spreads
out along lines with a vertical-to-horizontal slope of 2:1.
Example 5.1
A flexible rectangular area measures 2.5 m 3 5 m in plan. It supports a load of
150 kN> m2.
Determine the vertical stress increase due to the load at a depth of 6.25 m below the
center of the rectangular area.
Solution
Refer to Figure 5.4. For this case,
2.5
5 1.25 m
2
5
L1 5 L2 5 5 2.5 m
2
B1 5 B2 5
232 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
From Eqs. (5.7) and (5.8),
B1
B2
1.25
5
5
5 0.2
z
z
6.25
L1
L2
2.5
n5
5
5
5 0.4
z
z
6.25
m5
From Table 5.2, for m 5 0.2 and n 5 0.4, the value of I 5 0.0328. Thus,
Ds 5 qo (4I) 5 (150) (4) (0.0328) 5 19.68 kN , m2
Alternate Solution
From Eq. (5.10),
Ds 5 qoIc
L
5
m1 5 5
52
B
2.5
z
6.25
n1 5
5
55
B
2.5
a b
a b
2
2
From Table 5.3, for m1 5 2 and n1 5 5, the value of Ic 5 0.131. Thus,
Ds 5 (150) (0.131) 5 19.65 kN , m2
5.5
■
Average Vertical Stress Increase
Due to a Rectangularly Loaded Area
In Section 5.4, the vertical stress increase below the corner of a uniformly loaded rectangular area was given as
Ds 5 qoI
In many cases, one must find the average stress increase, Dsav , below the corner of a uniformly loaded rectangular area with limits of z 5 0 to z 5 H, as shown in Figure 5.6. This
can be evaluated as
H
Dsav 5
1
(qoI) dz 5 qoIa
H 30
(5.15)
where
Ia 5 f(m2, n2 )
B
m2 5
H
(5.16)
(5.17)
5.5 Average Vertical Stress Increase Due to a Rectangularly Loaded Area
233
qo /unit area
av
z
Section
of loaded
area
dz
H
A
(a)
z
(c)
B
Plan of
loaded
area
L
A
(b)
Figure 5.6 Average vertical stress increase due to a rectangularly loaded flexible area
and
n2 5
L
H
(5.18)
The variation of Ia with m2 and n2 is shown in Figure 5.7, as proposed by Griffiths (1984).
In estimating the consolidation settlement under a foundation, it may be required to
determine the average vertical stress increase in only a given layer—that is, between
z 5 H1 and z 5 H2 , as shown in Figure 5.8. This can be done as (Griffiths, 1984)
Dsav(H2>H1) 5 qo B
H2Ia(H2) 2 H1Ia(H1)
H2 2 H1
R
where
Dsav(H2>H1) 5 average stress increase immediately below the corner of a uniformly
loaded rectangular area between depths z H1 and z H2
Ia(H2) 5 Ia for z 5 0 to z 5 H2 5 f¢ m2 5
B
L
, n2 5
≤
H2
H2
Ia(H1) 5 Ia for z 5 0 to z 5 H1 5 f¢m2 5
B
L
, n2 5
≤
H1
H1
(5.19)
234 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
0.26
n2 2.0
1.0
0.8
0.6
0.5
0.4
0.24
0.22
0.20
0.18
0.3
0.16
0.14
Ia
0.2
0.12
0.1
0.1
0.08
0.06
0.04
0.02
0.00
0.1
0.2
0.3 0.4 0.50.6 0.8 1.0
2
3
4
5 6 7 8 9 10
m2
Figure 5.7 Griffiths’ influence factor Ia
qo /unit area
Section
H1
H2
av(H2/H1)
z
A
A
z
B
L
Plan
A, A
Figure 5.8 Average pressure increase between z 5 H1 and z 5 H2 below the
corner of a uniformly loaded rectangular area
5.5 Average Vertical Stress Increase Due to a Rectangularly Loaded Area
235
Example 5.2
Refer to Figure 5.9. Determine the average stress increase below the center of the
loaded area between z 5 3 m to z 5 5 m (that is, between points A and A9).
Solution
Refer to Figure 5.9. The loaded area can be divided into four rectangular areas, each
measuring 1.5 m 3 1.5 m (L 3 B). Using Eq. (5.19), the average stress increase
(between the required depths) below the corner of each rectangular area can be given as
Dsav (H2>H1) 5 qo B
H2Ia(H2) 2 H1Ia(H1)
H2 2 H1
R 5 100 B
(5)Ia(H2) 2 (3)Ia(H1)
523
R
For Ia(H2):
B
1.5
5
5 0.3
H2
5
L
1.5
n2 5
5
5 0.3
H2
5
m2 5
Referring to Figure 5.7, for m2 5 0.3 and n2 5 0.3, Ia(H2) 5 0.126. For Ia(H1):
B
1.5
5
5 0.5
H1
3
L
1.5
n2 5
5
5 0.5
H1
3
m2 5
qo 100 kN/m2
1.5 m
1.5 m
z
3m
5m
Section
A
A
B
A, A
3m
Plan
L
3m
Figure 5.9 Determination of average increase
in stress below a rectangular area
236 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Referring to Figure 5.7, Ia(H1) 5 0.175, so
Dsav (H2>H1) 5 100 B
(5) (0.126) 2 (3) (0.175)
R 5 5.25 kN , m2
523
The stress increase between z 5 3 m to z 5 5 m below the center of the loaded area is
equal to
4Dsav (H2>H1) 5 (4) (5.25) 5 21 kN , m2
5.6
■
Stress Increase under an Embankment
Figure 5.10 shows the cross section of an embankment of height H. For this two-dimensional
loading condition, the vertical stress increase may be expressed as
Ds 5
qo
B1 1 B2
B1
B¢
≤ (a1 1 a2 ) 2
(a ) R
p
B2
B2 2
(5.20)
where
qo 5 gH
g 5 unit weight of the embankment soil
H 5 height of the embankment
a1 5 tan21 ¢
B1 1 B2
B1
≤ 2 tan21 ¢ ≤
z
z
(5.21)
a2 5 tan21 ¢
B1
≤
z
(5.22)
(Note that a1 and a2 are in radians.)
B2
B1
qo H
H
1
2
z
Figure 5.10 Embankment
loading
5.6 Stress Increase under an Embankment
0.50
237
3.0
2.0
1.6
1.4
1.2
1.0
0.9
0.8
0.45
0.40
0.7
0.35
0.6
0.30
I
0.5
0.4
0.25
0.3
0.20
0.2
0.15
0.10
0.1
0.05
B1 /z = 0
0
0.01
0.1
1.0
B2 /z
Figure 5.11 Influence value Ir for
embankment loading (After Osterberg,
1957) (Osterberg, J. O. (1957). “Influence
Values for Vertical Stresses in Semi-Infinite
Mass Due to Embankment Loading,”
Proceedings, Fourth International
Conference on Soil Mechanics and
10 Foundation Engineering, London, Vol. 1.
pp. 393–396. With permission from ASCE.)
For a detailed derivation of Eq. (5.20), see Das (2008). A simplified form of the
equation is
Ds 5 qoIr
(5.23)
where Ir 5 a function of B1>z and B2>z.
The variation of Ir with B1>z and B2>z is shown in Figure 5.11. An application of
this diagram is given in Example 5.3.
Example 5.3
An embankment is shown in Figure 5.12a. Determine the stress increase under the
embankment at points A 1 and A 2 .
Solution
We have
gH 5 (17.5) (7) 5 122.5 kN>m2
238 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Stress Increase at A1
The left side of Figure 5.12b indicates that B1 5 2.5 m and B2 5 14 m, so
B1
2.5
5 0.5
5
z
5
and
B2
14
5 2.8
5
z
5
14 m
5m
14 m
H7m
17.5 kN/m3
5m
11.5 m
16.5 m
5m
5m
A2
A1
(a)
2.5 m
2.5 m
14 m
14 m
qo qo 122.5 122.5
kN/m2 kN/m2
5m
(1)
(2)
A2
A1
(b)
14 m
5m
14 m
qo (7 m)
qo (2.5 m)
3
(17.5 kN/m3) (17.5 kN/m )
2
122.5 kN/m2
43.75 kN/m
5m
(1)
A2
(2)
A2
qo (4.5 m)
(17.5 kN/m3)
78.75 kN/m2
9m
(3)
(c)
Figure 5.12 Stress increase due to embankment loading
A2
5.6 Stress Increase under an Embankment
239
According to Figure 5.11, in this case Ir 5 0.445. Because the two sides in Figure 5.12b
are symmetrical, the value of Ir for the right side will also be 0.445, so
Ds 5 Ds(1) 1 Ds(2) 5 qo 3I (left
r side) 1 I (right
r side) 4
5 122.530.445 1 0.4454 5 109.03 kN , m2
Stress Increase at A2
In Figure 5.12c, for the left side, B2 5 5 m and B1 5 0, so
B2
5
5 51
z
5
and
B1
0
5 50
z
5
According to Figure 5.11, for these values of B2>z and B1>z, I r 5 0.24; hence,
Ds(1) 5 43.75(0.24) 5 10.5 kN>m2
For the middle section,
B2
14
5
5 2.8
z
5
and
B1
14
5
5 2.8
z
5
Thus, I r 5 0.495, so
Ds(2) 5 0.495(122.5) 5 60.64 kN>m2
For the right side,
B2
9
5 5 1.8
z
5
B1
0
5 50
z
5
and I r 5 0.335, so
Ds(3) 5 (78.75) (0.335) 5 26.38 kN>m2
The total stress increase at point A 2 is
Ds 5 Ds(1) 1 Ds(2) 2 Ds(3) 5 10.5 1 60.64 2 26.38 5 44.76 kN , m2
■
240 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
5.7
Westergaard’s Solution for Vertical Stress
Due to a Point Load
Boussinesq’s solution for stress distribution due to a point load was presented in Section 5.2.
The stress distribution due to various types of loading discussed in Sections 5.3 through 5.6
is based on integration of Boussinesq’s solution.
Westergaard (1938) has proposed a solution for the determination of the vertical
stress due to a point load P in an elastic solid medium in which there exist alternating
layers with thin rigid reinforcements (Figure 5.13a). This type of assumption may be an
idealization of a clay layer with thin seams of sand. For such an assumption, the vertical stress increase at a point A (Figure 5.13b) can be given as
Ds 5
Ph
3>2
1
d
2
2pz h 1 (r>z)
2
c
2
(5.24)
P
Thin rigid reinforcement
s = Poisson s ratio of soil between the rigid layers
(a)
P
x
r
z
y
A
z
(b)
Figure 5.13 Westergaard’s solution
for vertical stress due to a point load
5.8 Stress Distribution for Westergaard Material
241
Table 5.4 Variation of I1 [Eq. (5.27)].
I1
r/z
s ⴝ 0
s ⴝ 0.2
s ⴝ 0.4
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.5
2.0
2.5
3.0
4.0
5.0
0.3183
0.3090
0.2836
0.2483
0.2099
0.1733
0.1411
0.1143
0.0925
0.0751
0.0613
0.0247
0.0118
0.0064
0.0038
0.0017
0.0009
0.4244
0.4080
0.3646
0.3074
0.2491
0.1973
0.1547
0.1212
0.0953
0.0756
0.0605
0.0229
0.0107
0.0057
0.0034
0.0015
0.0008
0.9550
0.8750
0.6916
0.4997
0.3480
0.2416
0.1700
0.1221
0.0897
0.0673
0.0516
0.0173
0.0076
0.0040
0.0023
0.0010
0.0005
where
1 2 2ms
Ä 2 2 2ms
s Poisson’s ratio of the solid between the rigid reinforcements
r 5 !x2 1 y2
Equation (5.24) can be rewritten as
h5
Ds 5 a
P
bI1
z2
(5.25)
(5.26)
where
I1 5
23>2
1
r 2
c
a
b
1
1d
2ph2 hz
(5.27)
Table 5.4 gives the variation of I1 with s.
5.8
Stress Distribution for Westergaard Material
Stress Due to a Circularly Loaded Area
Referring to Figure 5.2, if the circular area is located on a Westergaard-type material, the
increase in vertical stress, , at a point located at a depth z immediately below the center of the area can be given as
242 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Ds 5 qo μ 1 2
h
B 2 1>2
ch 1 a b d
2z
2
∂
(5.28)
The term has been defined in Eq. (5.25). The variations of /qo with B/2z and s 0
are given in Table 5.5.
Stress Due to a Uniformly Loaded Flexible Rectangular Area
Refer to Figure 5.3. If the flexible rectangular area is located on a Westergaard-type material, the stress increase at point A can be given as
Ds 5
qo
1
1
1
ccot21 h2 a 2 1 2 b 1 h4 a 2 2 b d
2p
Ä
m
n
mn
where
m5
B
z
n5
L
z
Table 5.5 Variation of /qo with B/2z
and s 0 [Eq. (5.28)]
B/2z
⌬/qo
0.00
0.25
0.33
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
4.00
5.00
6.00
7.00
8.00
9.00
10.00
0.0
0.0572
0.0938
0.1835
0.3140
0.4227
0.5076
0.5736
0.6254
0.6667
0.7002
0.7278
0.7510
0.7706
0.8259
0.8600
0.8830
0.8995
0.9120
0.9217
0.9295
(5.29a)
5.9 Elastic Settlement of Foundations on Saturated Clay (s 0.5)
243
Table 5.6. Variation of Iw with m and n (s 0).
n
m
0.1
0.2
0.4
0.5
0.6
1.0
2.0
5.0
10.0
0.1
0.2
0.4
0.5
0.6
1.0
2.0
5.0
10.0
0.0031
0.0061
0.0110
0.0129
0.0144
0.0183
0.0211
0.0221
0.0223
0.0061
0.0118
0.0214
0.0251
0.0282
0.0357
0.0413
0.0435
0.0438
0.0110
0.0214
0.0390
0.0459
0.0516
0.0658
0.0768
0.0811
0.0817
0.0129
0.0251
0.0459
0.0541
0.0610
0.0781
0.0916
0.0969
0.0977
0.0144
0.0282
0.0516
0.0610
0.0687
0.0886
0.1044
0.1107
0.1117
0.0182
0.0357
0.0658
0.0781
0.0886
0.1161
0.1398
0.1499
0.1515
0.0211
0.0413
0.0768
0.0916
0.1044
0.1398
0.1743
0.1916
0.1948
0.0211
0.0434
0.0811
0.0969
0.1107
0.1491
0.1916
0.2184
0.2250
0.0223
0.0438
0.0847
0.0977
0.1117
0.1515
0.1948
0.2250
0.2341
or
Ds
1
1
1
1
5
ccot 21 h2 a 2 1 2 b 1 h4 a 2 2 b d 5 Iw
qo
2p
Å
m
n
mn
(5.29b)
Table 5.6 gives the variation of Iw with m and n (for s 0).
Example 5.4
Solve Example 5.1 using Eq. (5.29). Assume s 0.
Solution
From Example 5.1
m 0.2
n 0.4
qo(4Iw)
From Table 5.6, for m 0.2 and n 0.4, the value of Iw ⬇ 0.0214. So
(150)(4 0.0214) 12.84 kN/m2
■
Elastic Settlement
5.9
Elastic Settlement of Foundations on Saturated
Clay (␮s 0.5)
Janbu et al. (1956) proposed an equation for evaluating the average settlement of flexible
foundations on saturated clay soils (Poisson’s ratio, s 0.5). For the notation used in
Figure 5.14, this equation is
244 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Se 5 A1A2
qoB
Es
(5.30)
where A1 is a function of H/B and L/B and A2 is a function of Df /B.
Christian and Carrier (1978) modified the values of A1 and A2 to some extent as presented in Figure 5.14.
qo
B
Df
H
1.0
A2 0.9
0.8
0
5
10
Df /B
15
20
2.0
L /B L /B 10
1.5
5
A1 1.0
2
Square
Circle
0.5
0
0.1
1
10
H/B
100
1000
Figure 5.14 Values of A1 and A2 for elastic settlement calculation—Eq. (5.30) (After
Christian and Carrier, 1978) (Christian, J. T. and and Carrier, W. D. (1978). “Janbu,
Bjerrum and Kjaernsli's chart reinterpreted,” Canadian Geotechnical Journal, Vol. 15,
pp. 123–128. © 2008 NRC Canada or its licensors. Reproduced with permission.)
5.10 Settlement Based on the Theory of Elasticity
245
Table 5.7 Range of for Clay [Eq. (5.31)]a
␤
Plasticity
Index
OCR 1
OCR 2
OCR 3
OCR 4
OCR 5
30
30 to 50
50
1500–600
600–300
300–150
1380–500
550–270
270–120
1200–580
580–220
220–100
950–380
380–180
180–90
730–300
300–150
150–75
a
Interpolated from Duncan and Buchignani, (1976)
The modulus of elasticity (Es) for clays can, in general, be given as
Es cu
(5.31)
where cu undrained shear strength.
The parameter is primarily a function of the plasticity index and overconsolidation
ratio. Table 5.7 provides a general range for based on that proposed by Duncan
and Buchignani (1976). In any case, proper judgment should be used in selecting the
magnitude of .
5.10
Settlement Based on the Theory of Elasticity
The elastic settlement of a shallow foundation can be estimated by using the theory of
elasticity. From Hooke’s law, as applied to Figure 5.15, we obtain
H
H
1
Se 5 3 ezdz 5
(Dsz 2 ms Dsx 2 ms Dsy )dz
Es 30
0
Load qo /unit area
z
y
H
dz
x
y
z
Incompressible
layer
Figure 5.15 Elastic settlement of shallow
foundation
(5.32)
246
Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
where
Se 5 elastic settlement
Es 5 modulus of elasticity of soil
H 5 thickness of the soil layer
ms 5 Poisson’s ratio of the soil
Dsx , Dsy , Dsz 5 stress increase due to the net applied foundation load in the x, y, and
z directions, respectively
Theoretically, if the foundation is perfectly flexible (see Figure 5.16 and Bowles,
1987), the settlement may be expressed as
Se 5 qo (aBr )
1 2 m2s
IsIf
Es
(5.33)
where
qo 5 net applied pressure on the foundation
ms 5 Poisson’s ratio of soil
Es 5 average modulus of elasticity of the soil under the foundation, measured from
z 5 0 to about z 5 5B
B r 5 B>2 for center of foundation
5 B for corner of foundation
Is 5 shape factor (Steinbrenner, 1934)
5 F1 1
F1 5
Rigid
foundation
settlement
(5.34)
1
(A 1 A 1 )
p 0
Foundation B L
z
1 2 2ms
F
1 2 ms 2
qo
(5.35)
Df
Flexible
foundation
settlement H
s Poisson s ratio
Es Modulus of elasticity
Soil
Rock
Figure 5.16 Elastic settlement of
flexible and rigid foundations
5.10 Settlement Based on the Theory of Elasticity
F2 5
nr
tan21A 2
2p
A 0 5 mrln
(5.36)
¢1 1 "mr2 1 1≤"mr2 1 nr2
A2 5
(5.37)
mr ¢1 1 "mr 1 nr 1 1≤
2
A 1 5 ln
247
2
¢ mr 1 "mr2 1 1≤"1 1 nr2
(5.38)
mr 1 "mr2 1 nr2 1 1
mr
nr"mr2 1 nr2 1 1
If 5 depth factor (Fox, 1948) 5 f¢
(5.39)
Df
B
, ms , and
L
≤
B
(5.40)
a 5 a factor that depends on the location on the
foundation where settlement is being calculated
To calculate settlement at the center of the foundation, we use
a54
L
mr 5
B
and
H
nr 5
¢
B
≤
2
To calculate settlement at a corner of the foundation,
a51
L
mr 5
B
and
nr 5
H
B
The variations of F1 and F2 [see Eqs. (5.35) and (5.36)] with mr and nr are given
in Tables 5.8 and 5.9. Also, the variation of If with Df>B (for ms 5 0.3, 0.4, and 0.5) is
given in Table 5.10. These values are also given in more detailed form by Bowles
(1987).
The elastic settlement of a rigid foundation can be estimated as
Se(rigid) < 0.93Se(flexible, center)
(5.41)
248 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Table 5.8 Variation of F1 with m9 and n9
mⴕ
nⴕ
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
9.25
9.50
9.75
10.00
20.00
50.00
100.00
1.0
1.2
1.4
1.6
1.8
2.0
2.5
3.0
3.5
4.0
0.014
0.049
0.095
0.142
0.186
0.224
0.257
0.285
0.309
0.330
0.348
0.363
0.376
0.388
0.399
0.408
0.417
0.424
0.431
0.437
0.443
0.448
0.453
0.457
0.461
0.465
0.468
0.471
0.474
0.477
0.480
0.482
0.485
0.487
0.489
0.491
0.493
0.495
0.496
0.498
0.529
0.548
0.555
0.013
0.046
0.090
0.138
0.183
0.224
0.259
0.290
0.317
0.341
0.361
0.379
0.394
0.408
0.420
0.431
0.440
0.450
0.458
0.465
0.472
0.478
0.483
0.489
0.493
0.498
0.502
0.506
0.509
0.513
0.516
0.519
0.522
0.524
0.527
0.529
0.531
0.533
0.536
0.537
0.575
0.598
0.605
0.012
0.044
0.087
0.134
0.179
0.222
0.259
0.292
0.321
0.347
0.369
0.389
0.406
0.422
0.436
0.448
0.458
0.469
0.478
0.487
0.494
0.501
0.508
0.514
0.519
0.524
0.529
0.533
0.538
0.541
0.545
0.549
0.552
0.555
0.558
0.560
0.563
0.565
0.568
0.570
0.614
0.640
0.649
0.011
0.042
0.084
0.130
0.176
0.219
0.258
0.292
0.323
0.350
0.374
0.396
0.415
0.431
0.447
0.460
0.472
0.484
0.494
0.503
0.512
0.520
0.527
0.534
0.540
0.546
0.551
0.556
0.561
0.565
0.569
0.573
0.577
0.580
0.583
0.587
0.589
0.592
0.595
0.597
0.647
0.678
0.688
0.011
0.041
0.082
0.127
0.173
0.216
0.255
0.291
0.323
0.351
0.377
0.400
0.420
0.438
0.454
0.469
0.481
0.495
0.506
0.516
0.526
0.534
0.542
0.550
0.557
0.563
0.569
0.575
0.580
0.585
0.589
0.594
0.598
0.601
0.605
0.609
0.612
0.615
0.618
0.621
0.677
0.711
0.722
0.011
0.040
0.080
0.125
0.170
0.213
0.253
0.289
0.322
0.351
0.378
0.402
0.423
0.442
0.460
0.476
0.484
0.503
0.515
0.526
0.537
0.546
0.555
0.563
0.570
0.577
0.584
0.590
0.596
0.601
0.606
0.611
0.615
0.619
0.623
0.627
0.631
0.634
0.638
0.641
0.702
0.740
0.753
0.010
0.038
0.077
0.121
0.165
0.207
0.247
0.284
0.317
0.348
0.377
0.402
0.426
0.447
0.467
0.484
0.495
0.516
0.530
0.543
0.555
0.566
0.576
0.585
0.594
0.603
0.610
0.618
0.625
0.631
0.637
0.643
0.648
0.653
0.658
0.663
0.667
0.671
0.675
0.679
0.756
0.803
0.819
0.010
0.038
0.076
0.118
0.161
0.203
0.242
0.279
0.313
0.344
0.373
0.400
0.424
0.447
0.458
0.487
0.514
0.521
0.536
0.551
0.564
0.576
0.588
0.598
0.609
0.618
0.627
0.635
0.643
0.650
0.658
0.664
0.670
0.676
0.682
0.687
0.693
0.697
0.702
0.707
0.797
0.853
0.872
0.010
0.037
0.074
0.116
0.158
0.199
0.238
0.275
0.308
0.340
0.369
0.396
0.421
0.444
0.466
0.486
0.515
0.522
0.539
0.554
0.568
0.581
0.594
0.606
0.617
0.627
0.637
0.646
0.655
0.663
0.671
0.678
0.685
0.692
0.698
0.705
0.710
0.716
0.721
0.726
0.830
0.895
0.918
0.010
0.037
0.074
0.115
0.157
0.197
0.235
0.271
0.305
0.336
0.365
0.392
0.418
0.441
0.464
0.484
0.515
0.522
0.539
0.554
0.569
0.584
0.597
0.609
0.621
0.632
0.643
0.653
0.662
0.671
0.680
0.688
0.695
0.703
0.710
0.716
0.723
0.719
0.735
0.740
0.858
0.931
0.956
5.10 Settlement Based on the Theory of Elasticity
249
Table 5.8 (Continued)
mⴕ
nⴕ
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
9.25
9.50
9.75
10.00
20.00
50.00
100.00
4.5
5.0
6.0
0.010
0.036
0.073
0.114
0.155
0.195
0.233
0.269
0.302
0.333
0.362
0.389
0.415
0.438
0.461
0.482
0.516
0.520
0.537
0.554
0.569
0.584
0.597
0.611
0.623
0.635
0.646
0.656
0.666
0.676
0.685
0.694
0.702
0.710
0.717
0.725
0.731
0.738
0.744
0.750
0.878
0.962
0.990
0.010
0.036
0.073
0.113
0.154
0.194
0.232
0.267
0.300
0.331
0.359
0.386
0.412
0.435
0.458
0.479
0.496
0.517
0.535
0.552
0.568
0.583
0.597
0.610
0.623
0.635
0.647
0.658
0.669
0.679
0.688
0.697
0.706
0.714
0.722
0.730
0.737
0.744
0.751
0.758
0.896
0.989
1.020
0.010
0.036
0.072
0.112
0.153
0.192
0.229
0.264
0.296
0.327
0.355
0.382
0.407
0.430
0.453
0.474
0.484
0.513
0.530
0.548
0.564
0.579
0.594
0.608
0.621
0.634
0.646
0.658
0.669
0.680
0.690
0.700
0.710
0.719
0.727
0.736
0.744
0.752
0.759
0.766
0.925
1.034
1.072
7.0
8.0
9.0
10.0
25.0
50.0
100.0
0.010
0.036
0.072
0.112
0.152
0.191
0.228
0.262
0.294
0.324
0.352
0.378
0.403
0.427
0.449
0.470
0.473
0.508
0.526
0.543
0.560
0.575
0.590
0.604
0.618
0.631
0.644
0.656
0.668
0.679
0.689
0.700
0.710
0.719
0.728
0.737
0.746
0.754
0.762
0.770
0.945
1.070
1.114
0.010
0.036
0.072
0.112
0.152
0.190
0.227
0.261
0.293
0.322
0.350
0.376
0.401
0.424
0.446
0.466
0.471
0.505
0.523
0.540
0.556
0.571
0.586
0.601
0.615
0.628
0.641
0.653
0.665
0.676
0.687
0.698
0.708
0.718
0.727
0.736
0.745
0.754
0.762
0.770
0.959
1.100
1.150
0.010
0.036
0.072
0.111
0.151
0.190
0.226
0.260
0.291
0.321
0.348
0.374
0.399
0.421
0.443
0.464
0.471
0.502
0.519
0.536
0.553
0.568
0.583
0.598
0.611
0.625
0.637
0.650
0.662
0.673
0.684
0.695
0.705
0.715
0.725
0.735
0.744
0.753
0.761
0.770
0.969
1.125
1.182
0.010
0.036
0.071
0.111
0.151
0.189
0.225
0.259
0.291
0.320
0.347
0.373
0.397
0.420
0.441
0.462
0.470
0.499
0.517
0.534
0.550
0.585
0.580
0.595
0.608
0.622
0.634
0.647
0.659
0.670
0.681
0.692
0.703
0.713
0.723
0.732
0.742
0.751
0.759
0.768
0.977
1.146
1.209
0.010
0.036
0.071
0.110
0.150
0.188
0.223
0.257
0.287
0.316
0.343
0.368
0.391
0.413
0.433
0.453
0.468
0.489
0.506
0.522
0.537
0.551
0.565
0.579
0.592
0.605
0.617
0.628
0.640
0.651
0.661
0.672
0.682
0.692
0.701
0.710
0.719
0.728
0.737
0.745
0.982
1.265
1.408
0.010
0.036
0.071
0.110
0.150
0.188
0.223
0.256
0.287
0.315
0.342
0.367
0.390
0.412
0.432
0.451
0.462
0.487
0.504
0.519
0.534
0.549
0.583
0.576
0.589
0.601
0.613
0.624
0.635
0.646
0.656
0.666
0.676
0.686
0.695
0.704
0.713
0.721
0.729
0.738
0.965
1.279
1.489
0.010
0.036
0.071
0.110
0.150
0.188
0.223
0.256
0.287
0.315
0.342
0.367
0.390
0.411
0.432
0.451
0.460
0.487
0.503
0.519
0.534
0.548
0.562
0.575
0.588
0.600
0.612
0.623
0.634
0.645
0.655
0.665
0.675
0.684
0.693
0.702
0.711
0.719
0.727
0.735
0.957
1.261
1.499
250 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Table 5.9 Variation of F2 with m9 and n9
mⴕ
nⴕ
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
9.25
9.50
9.75
10.00
20.00
50.00
100.00
1.0
1.2
1.4
1.6
1.8
2.0
2.5
3.0
0.049
0.074
0.083
0.083
0.080
0.075
0.069
0.064
0.059
0.055
0.051
0.048
0.045
0.042
0.040
0.037
0.036
0.034
0.032
0.031
0.029
0.028
0.027
0.026
0.025
0.024
0.023
0.022
0.022
0.021
0.020
0.020
0.019
0.018
0.018
0.017
0.017
0.017
0.016
0.016
0.008
0.003
0.002
0.050
0.077
0.089
0.091
0.089
0.084
0.079
0.074
0.069
0.064
0.060
0.056
0.053
0.050
0.047
0.044
0.042
0.040
0.038
0.036
0.035
0.033
0.032
0.031
0.030
0.029
0.028
0.027
0.026
0.025
0.024
0.023
0.023
0.022
0.021
0.021
0.020
0.020
0.019
0.019
0.010
0.004
0.002
0.051
0.080
0.093
0.098
0.096
0.093
0.088
0.083
0.077
0.073
0.068
0.064
0.060
0.057
0.054
0.051
0.049
0.046
0.044
0.042
0.040
0.039
0.037
0.036
0.034
0.033
0.032
0.031
0.030
0.029
0.028
0.027
0.026
0.026
0.025
0.024
0.024
0.023
0.023
0.022
0.011
0.004
0.002
0.051
0.081
0.097
0.102
0.102
0.099
0.095
0.090
0.085
0.080
0.076
0.071
0.067
0.064
0.060
0.057
0.055
0.052
0.050
0.048
0.046
0.044
0.042
0.040
0.039
0.038
0.036
0.035
0.034
0.033
0.032
0.031
0.030
0.029
0.028
0.028
0.027
0.026
0.026
0.025
0.013
0.005
0.003
0.051
0.083
0.099
0.106
0.107
0.105
0.101
0.097
0.092
0.087
0.082
0.078
0.074
0.070
0.067
0.063
0.061
0.058
0.055
0.053
0.051
0.049
0.047
0.045
0.044
0.042
0.041
0.039
0.038
0.037
0.036
0.035
0.034
0.033
0.032
0.031
0.030
0.029
0.029
0.028
0.014
0.006
0.003
0.052
0.084
0.101
0.109
0.111
0.110
0.107
0.102
0.098
0.093
0.089
0.084
0.080
0.076
0.073
0.069
0.066
0.063
0.061
0.058
0.056
0.054
0.052
0.050
0.048
0.046
0.045
0.043
0.042
0.041
0.039
0.038
0.037
0.036
0.035
0.034
0.033
0.033
0.032
0.031
0.016
0.006
0.003
0.052
0.086
0.104
0.114
0.118
0.118
0.117
0.114
0.110
0.106
0.102
0.097
0.093
0.089
0.086
0.082
0.079
0.076
0.073
0.070
0.067
0.065
0.063
0.060
0.058
0.056
0.055
0.053
0.051
0.050
0.048
0.047
0.046
0.045
0.043
0.042
0.041
0.040
0.039
0.038
0.020
0.008
0.004
0.052
0.086
0.106
0.117
0.122
0.124
0.123
0.121
0.119
0.115
0.111
0.108
0.104
0.100
0.096
0.093
0.090
0.086
0.083
0.080
0.078
0.075
0.073
0.070
0.068
0.066
0.064
0.062
0.060
0.059
0.057
0.055
0.054
0.053
0.051
0.050
0.049
0.048
0.047
0.046
0.024
0.010
0.005
3.5
4.0
0.052
0.0878
0.107
0.119
0.125
0.128
0.128
0.127
0.125
0.122
0.119
0.116
0.112
0.109
0.105
0.102
0.099
0.096
0.093
0.090
0.087
0.084
0.082
0.079
0.077
0.075
0.073
0.071
0.069
0.067
0.065
0.063
0.062
0.060
0.059
0.057
0.056
0.055
0.054
0.052
0.027
0.011
0.006
0.052
0.087
0.108
0.120
0.127
0.130
0.131
0.131
0.130
0.127
0.125
0.122
0.119
0.116
0.113
0.110
0.107
0.104
0.101
0.098
0.095
0.092
0.090
0.087
0.085
0.083
0.080
0.078
0.076
0.074
0.072
0.071
0.069
0.067
0.066
0.064
0.063
0.061
0.060
0.059
0.031
0.013
0.006
5.10 Settlement Based on the Theory of Elasticity
251
Table 5.9 (Continued)
mⴕ
nⴕ
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.75
7.00
7.25
7.50
7.75
8.00
8.25
8.50
8.75
9.00
9.25
9.50
9.75
10.00
20.00
50.00
100.00
4.5
5.0
6.0
7.0
8.0
9.0
10.0
25.0
50.0
100.0
0.053
0.087
0.109
0.121
0.128
0.132
0.134
0.134
0.133
0.132
0.130
0.127
0.125
0.122
0.119
0.116
0.113
0.110
0.107
0.105
0.102
0.099
0.097
0.094
0.092
0.090
0.087
0.085
0.083
0.081
0.079
0.077
0.076
0.074
0.072
0.071
0.069
0.068
0.066
0.065
0.035
0.014
0.007
0.053
0.087
0.109
0.122
0.130
0.134
0.136
0.136
0.136
0.135
0.133
0.131
0.129
0.126
0.124
0.121
0.119
0.116
0.113
0.111
0.108
0.106
0.103
0.101
0.098
0.096
0.094
0.092
0.090
0.088
0.086
0.084
0.082
0.080
0.078
0.077
0.075
0.074
0.072
0.071
0.039
0.016
0.008
0.053
0.088
0.109
0.123
0.131
0.136
0.138
0.139
0.140
0.139
0.138
0.137
0.135
0.133
0.131
0.129
0.127
0.125
0.123
0.120
0.118
0.116
0.113
0.111
0.109
0.107
0.105
0.103
0.101
0.099
0.097
0.095
0.093
0.091
0.089
0.088
0.086
0.085
0.083
0.082
0.046
0.019
0.010
0.053
0.088
0.110
0.123
0.132
0.137
0.140
0.141
0.142
0.142
0.142
0.141
0.140
0.138
0.137
0.135
0.133
0.131
0.130
0.128
0.126
0.124
0.122
0.120
0.118
0.116
0.114
0.112
0.110
0.108
0.106
0.104
0.102
0.101
0.099
0.097
0.096
0.094
0.092
0.091
0.053
0.022
0.011
0.053
0.088
0.110
0.124
0.132
0.138
0.141
0.143
0.144
0.144
0.144
0.144
0.143
0.142
0.141
0.139
0.138
0.136
0.135
0.133
0.131
0.130
0.128
0.126
0.124
0.122
0.121
0.119
0.117
0.115
0.114
0.112
0.110
0.108
0.107
0.105
0.104
0.102
0.100
0.099
0.059
0.025
0.013
0.053
0.088
0.110
0.124
0.133
0.138
0.142
0.144
0.145
0.146
0.146
0.145
0.145
0.144
0.143
0.142
0.141
0.140
0.139
0.137
0.136
0.134
0.133
0.131
0.129
0.128
0.126
0.125
0.123
0.121
0.120
0.118
0.117
0.115
0.114
0.112
0.110
0.109
0.107
0.106
0.065
0.028
0.014
0.053
0.088
0.110
0.124
0.133
0.139
0.142
0.145
0.146
0.147
0.147
0.147
0.147
0.146
0.145
0.145
0.144
0.143
0.142
0.140
0.139
0.138
0.136
0.135
0.134
0.132
0.131
0.129
0.128
0.126
0.125
0.124
0.122
0.121
0.119
0.118
0.116
0.115
0.113
0.112
0.071
0.031
0.016
0.053
0.088
0.111
0.125
0.134
0.140
0.144
0.147
0.149
0.151
0.152
0.152
0.153
0.153
0.154
0.154
0.154
0.154
0.154
0.154
0.154
0.154
0.154
0.153
0.153
0.153
0.153
0.152
0.152
0.152
0.151
0.151
0.150
0.150
0.150
0.149
0.149
0.148
0.148
0.147
0.124
0.071
0.039
0.053
0.088
0.111
0.125
0.134
0.140
0.144
0.147
0.150
0.151
0.152
0.153
0.154
0.155
0.155
0.155
0.156
0.156
0.156
0.156
0.156
0.156
0.157
0.157
0.157
0.157
0.157
0.157
0.157
0.156
0.156
0.156
0.156
0.156
0.156
0.156
0.156
0.156
0.156
0.156
0.148
0.113
0.071
0.053
0.088
0.111
0.125
0.134
0.140
0.145
0.148
0.150
0.151
0.153
0.154
0.154
0.155
0.155
0.156
0.156
0.156
0.157
0.157
0.157
0.157
0.157
0.157
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.158
0.156
0.142
0.113
252 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Table 5.10 Variation of If with Df /B, B/L, and s
B/L
␮s
Df/B
0.2
0.5
1.0
0.3
0.2
0.4
0.6
1.0
0.2
0.4
0.6
1.0
0.2
0.4
0.6
1.0
0.95
0.90
0.85
0.78
0.97
0.93
0.89
0.82
0.99
0.95
0.92
0.85
0.93
0.86
0.80
0.71
0.96
0.89
0.84
0.75
0.98
0.93
0.87
0.79
0.90
0.81
0.74
0.65
0.93
0.85
0.78
0.69
0.96
0.89
0.82
0.72
0.4
0.5
Due to the nonhomogeneous nature of soil deposits, the magnitude of Es may vary
with depth. For that reason, Bowles (1987) recommended using a weighted average of Es
in Eq. (5.33), or
gEs(i) Dz
(5.42)
Es 5
z
where
Es(i) 5 soil modulus of elasticity within a depthDz
z 5 H or 5B, whichever is smaller
Example 5.5
Refer to Figure 5.16 and consider a rigid square foundation 2.44 m 2.44 m in plan (Df 1.22 m) on a layer of normally consolidated sand. A rock layer is located at z 10.98 m.
The following is an approximation of the standard penetration number (N60) with z.
z (m)
N60
0–2.44
2.44–21
6.4–36
7
6.4
10.98
Given: s 0.3 and qo 167.7 kN/m2. Estimate the elastic settlement of the foundation. Use Eq. (2.29).
5.10 Settlement Based on the Theory of Elasticity
253
Solution
Calculation of Average Es
From Eq. (2.29),
Es ⬇ pa N60
B 2.44 m, pa ⬇ 100 kN/m , and
2
Given: H 10.98 m
⬇ 10 (since it is a normally consolidated clean sand).
5B.
The approximate variations of Es using Eq. (2.29) are given below.
z (m)
z (m)
N60
Es (kN/m2)
2.44
3.96
4.58
7
11
14
7000
11,000
14,000
0–2.44
2.44–6.4
6.4–10.98
Using Eq. (5.42),
Es 5
SEs(i) Dz
z
5
(7,000) (2.44) 1 (11,000) (3.96) 1 (14,000) (4.58)
10.98
5 11,362 kN>m2
Calculation of Se below the Center of the Foundation [Eq. (5.33)]
Se 5 qo (aBr )
1 2 m2s
II
Es s f
2.44
5 1.22 m
2
a54
Br 5
L
51
B
H
10.98
nr 5
5
59
B
2.44
a b
a
b
2
2
mr 5
Is 5 F1 1
1 2 2ms
F
1 2 ms 2
From Table 5.8, F1 0.491. From Table 5.9, F2 0.017.
1 2 (2) (0.3)
d (0.017) 5 0.5007
1 2 0.3
For s 3, Df /B 1.22/2.44 0.5, and B/L 1, the value of If is about 0.78 (Table 5.10).
Is 5 0.491 1 c
Se 5 (167.7) (4 3 1.22) a
1 2 0.32
b (0.5007) (0.78) 5 0.0256 m < 25.6 mm
11,362
254 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Calculation of Se for Rigid Foundation
From Eq. (5.41),
Se(rigid) ⬇ 0.93Se(flexible, center) (25.6)(0.93) 23.81 mm ⬇ 24 mm
5.11
■
Improved Equation for Elastic Settlement
In 1999, Mayne and Poulos presented an improved formula for calculating the elastic
settlement of foundations. The formula takes into account the rigidity of the foundation,
the depth of embedment of the foundation, the increase in the modulus of elasticity of
the soil with depth, and the location of rigid layers at a limited depth. To use Mayne and
Poulos’s equation, one needs to determine the equivalent diameter Be of a rectangular
foundation, or
Be 5
4BL
Å p
(5.43)
where
B 5 width of foundation
L 5 length of foundation
For circular foundations,
Be 5 B
(5.44)
where B 5 diameter of foundation.
Figure 5.17 shows a foundation with an equivalent diameter Be located at a depth Df
below the ground surface. Let the thickness of the foundation be t and the modulus of elasticity of the foundation material be Ef . A rigid layer is located at a depth H below the
bottom of the foundation. The modulus of elasticity of the compressible soil layer can be
given as
Es 5 Eo 1 kz
(5.45)
qo
Be
Df
Ef
t
Compressible
soil layer
Es
s
Eo
Es
Es Eo kz
H
Rigid layer
Depth, z
Figure 5.17 Improved equation
for calculating elastic settlement:
general parameters
5.11 Improved Equation for Elastic Settlement
255
With the preceding parameters defined, the elastic settlement below the center of the
foundation is
Se 5
qoBeIGIFIE
a1 2 m2s b
Eo
(5.46)
where
IG 5 influence factor for the variation of Es with depth
Eo H
5 f¢b 5
, ≤
kBe Be
IF 5 foundation rigidity correction factor
IE 5 foundation embedment correction factor
Figure 5.18 shows the variation of IG with b 5 Eo>kBe and H>Be . The foundation rigidity
correction factor can be expressed as
IF 5
p
1
4
1
(5.47)
Ef
4.6 1 10 §
Eo 1
Be
k
2
¥¢
2t 3
≤
Be
Similarly, the embedment correction factor is
IE 5 1 2
1
Be
3.5 exp(1.22ms 2 0.4) ¢
1 1.6≤
Df
(5.48)
Figures 5.19 and 5.20 show the variation of IF and IE with terms expressed in Eqs. (5.47)
and (5.48).
1.0
10.0
30
5.0
0.8
2.0
1.0
0.4
0.5
IG
0.6
0.2
H/Be 0.2
0
0.01 2 4 6 0.1
1
E
kBo
e
10
100
Figure 5.18 Variation of IG with b
256 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
1.0
0.95
IF
0.9
0.85
KF (
Ef
3
)( )
2t
Be
Be
k
2
Flexibility factor
0.8
Eo 0.75
0.7
0.001 2 4 0.01
0.1
1.0
10.0
100
KF
Figure 5.19 Variation of
rigidity correction factor IF
with flexibility factor KF
[Eq. (5.47)]
1.0
0.95
IE
0.9
s = 0.5
0.4
0.85
0.3
0.2
0.8
0.1
0
0.75
0.7
0
5
10
Df
Be
15
20
Figure 5.20 Variation of
embedment correction factor IE
with Df>Be [Eq (5.48)]
Example 5.6
For a shallow foundation supported by a silty clay, as shown in Figure 5.17,
Length 5 L 5 3.05 m
Width 5 B 5 1.52 m
Depth of foundation 5 Df 5 1.52 m
5.11 Improved Equation for Elastic Settlement
257
Thickness of foundation 5 t 5 0.305 m
Load per unit area 5 qo 5 239.6 kN>m2
Ef 5 15.87 3 106 kN>m2
The silty clay soil has the following properties:
H 5 3.66 m
ms 5 0.3
Eo 5 9660 kN>m2
k 5 565.6 kN>m2>m
Estimate the elastic settlement of the foundation.
Solution
From Eq. (5.43), the equivalent diameter is
Be 5
(4) (1.52) (3.05)
4BL
5
5 2.43 m
p
p
Å
Å
so
b5
Eo
9660
5
5 7.02
kBe
(565.6) (2.43)
and
3.66
H
5
5 1.5
Be
2.43
From Figure 5.18, for b 5 7.02 and H>Be 5 1.5, the value of IG < 0.69. From Eq. (5.47),
IF 5
p
1
4
1
4.6 1 10 §
Ef
Eo 1
5
Be
k
2
p
1
4
¥¢
2t 3
≤
Be
1
4.6 1 10 D
5 0.785
15.87 3 106
9660 1 ¢
2.43
≤ (565.6)
2
TB
(2) (0.305) 3
R
2.43
From Eq. (5.48 ),
IE 5 1 2
1
3.5 exp(1.22ms 2 0.4) ¢
Be
1 1.6≤
Df
258 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
512
1
2.43
3.5 exp 3(1.22) (0.3) 2 0.44 ¢
1 1.6≤
1.52
5 0.908
From Eq. (5.46),
Se 5
qoBeIGIFIE
(1 2 m2s )
Eo
so, with qo 5 239.6 kN>m2, it follows that
Se 5
5.12
(239.6) (2.43) (0.69) (0.785) (0.908)
(1 2 0.32 ) 5 0.027 m 5 27 mm
(9660)
■
Settlement of Sandy Soil: Use of Strain Influence Factor
The settlement of granular soils can also be evaluated by the use of a semiempirical
strain influence factor proposed by Schmertmann et al. (1978). According to this method
(Figure 5.21), the settlement is
z2 I
z
Se 5 C1C2 (q 2 q) a Dz
E
0
s
(5.49)
where
Iz 5 strain influence factor
C1 5 a correction factor for the depth of foundation embedment 5 1 2 0.5
3q> (q 2 q)4
C2 5 a correction factor to account for creep in soil
5 1 1 0.2 log (time in years>0.1)
q 5 stress at the level of the foundation
q 5 gDf 5 effective stress at the base of the foundation
Es 5 modulus of elasticity of soil
The recommended variation of the strain influence factor Iz for square (L> B 5 1) or
circular foundations and for foundations with L> B $ 10 is shown in Figure 5.21. The Iz
diagrams for 1 , L> B , 10 can be interpolated.
Note that the maximum value of Iz [that is, Iz(m)] occurs at z z1 and then reduces to
zero at z z2. The maximum value of Iz can be calculated as
where
q# 2 q
Iz(m) 5 0.5 1 0.1
Ä qz(1)
r
q z(1) effective stress at a depth of z1 before construction of the foundation
(5.50)
259
5.12 Settlement of Sandy Soil: Use of Strain Influence Factor
q
Df
Iz (m)
Iz (m)
q = Df
Iz
0.1
0.2
Iz
qz(1)
B
z1 = 0.5B
qz(1)
z1 = B
z2 = 2B
L/B = 1
z
z
L/B ≥ 10
z2 = 4B
z
Figure 5.21 Variation of strain influence factor with depth and L/B
The following relations are suggested by Salgado (2008) for interpolation of Iz at
z 0, z1/B, and z2/B for rectangular foundations.
•
Iz at z 0
L
2 1b # 0.2
B
(5.51)
z1
L
5 0.5 1 0.0555a 2 1b # 1
B
B
(5.52)
z2
L
5 2 1 0.222a 2 1b # 4
B
B
(5.53)
Iz 5 0.1 1 0.0111a
•
•
Variation of z1/B for Iz(m)
Variation of z2/B
Schmertmann et al. (1978) suggested that
Es 2.5qc (for square foundation)
(5.54)
and
Es 3.5qc (for L/B
where qc cone penetration resistance.
10)
(5.55)
260 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
It appears reasonable to write (Terzaghi et al., 1996)
L
Es( rectangle) 5 a1 1 0.4 log bEs( square)
B
(5.56)
The procedure for calculating elastic settlement using Eq. (5.49) is given here
(Figure 5.22).
Step 1. Plot the foundation and the variation of Iz with depth to scale (Figure 5.22a).
Step 2. Using the correlation from standard penetration resistance (N60) or cone
penetration resistance (qc), plot the actual variation of Es with depth
(Figure 5.22b).
Step 3. Approximate the actual variation of Es into a number of layers of soil having a constant Es, such as Es(1), Es(2), . . . , Es(i), . . . Es(n) (Figure 5.22b).
Step 4. Divide the soil layer from z 5 0 to z 5 z2 into a number of layers by drawing
horizontal lines. The number of layers will depend on the break in continuity
in the Iz and Es diagrams.
Iz
Step 5. Prepare a table (such as Table 5.11) to obtain S Dz.
Es
Step 6. Calculate C1 and C2.
Step 7. Calculate Se from Eq. (5.49).
B
Df
z(1)
Es
Es(1)
Iz(1)
Step 4
z1
z(2)
Iz(2)
Es(2)
Step 3
Iz(3)
z2
z(i)
z(n)
Iz(i)
Es(i)
Step 1
Iz(n)
Depth, z
(a)
Es(n)
Depth, z
(b)
Figure 5.22 Procedure for calculation of Se using the strain influence factor
Step 2
5.12 Settlement of Sandy Soil: Use of Strain Influence Factor
Table 5.11 Calculation of S
Iz
Es
Dz
Layer
no.
Dz
Es
Iz at the middle
of the layer
1
2
Dz(1)
Dz(2)
Es(1)
Es(2)
Iz(1)
Iz(2)
(
(
(
(
i
Dz(i)
Es(i)
Iz(i)
(
(
(
(
n
Dz(n)
261
Es(n)
Iz(n)
Iz
Dz
Es
Iz(1)
Es(1)
Iz(i)
Es(i)
Dz1
Dzi
(
Iz(n)
Es(n)
S
Iz
Es
Dzn
Dz
Example 5.7
Consider a rectangular foundation 2 m 4 m in plan at a depth of 1.2 m in a sand deposit,
as shown in Figure 5.23a. Given: 17.5 kN/m3; q– 145 kN/m2, and the following
approximated variation of qc with z:
z (m)
qc (kN/m2)
0–0.5
0.5–2.5
2.5–5.0
2250
3430
2950
Estimate the elastic settlement of the foundation using the strain influence factor method.
Solution
From Eq. (5.52),
z1
L
4
5 0.5 1 0.0555a 2 1b 5 0.5 1 0.0555a 2 1b < 0.56
B
B
2
z1 (0.56)(2) 1.12 m
From Eq. (5.53),
z2
L
5 2 1 0.222a 2 1b 5 2 1 0.222(2 2 1) 5 2.22
B
B
z2 (2.22)(2) 4.44 m
262 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
From Eq. (5.51), at z 0,
Iz 5 0.1 1 0.0111a
L
4
2 1b 5 0.1 1 0.0111a 2 1b < 0.11
B
2
From Eq. (5.50),
q# 2 q
145 2 (1.2 3 17.5) 0.5
Iz(m) 5 0.5 1 0.1
5
0.5
1
0.1
c
d 5 0.675
Å q9z(1)
(1.2 1 1.12) (17.5)
The plot of Iz versus z is shown in Figure 5.23c. Again, from Eq. (5.56)
L
4
Es(rectangle) 5 a1 1 0.4log bEs(square) 5 c1 1 0.4loga b d (2.5 3 qc ) 5 2.8qc
B
2
Hence, the approximated variation of Es with z is as follows:
z (m)
qc (kN/m2)
Es (kN/m2)
0–0.5
0.5–2.5
2.5–5.0
2250
3430
2950
6300
9604
8260
The plot of Es versus z is shown in Figure 5.23b.
q = 145 kN/m2
1.2 m
B=2 m
L=4 m
= 17.5 kN/m3
0.5
0.675
0.11
6300
kN/m2
1.0
2.0
z
Es (kN/m2)
1
2
1.12
9604
kN/m2
3
2.5
3.0
8260
kN/m2
(a)
4
4.0
4.44
5.0
z (m)
(b)
Figure 5.23
z (m)
(c)
Iz
5.13 Settlement of Foundation on Sand Based on Standard Penetration Resistance
263
The soil layer is divided into four layers as shown in Figures 5.23b and 5.23c. Now
the following table can be prepared.
Layer no.
z (m)
Es (kN/m2)
Iz at middle
of layer
1
2
3
4
0.50
0.62
1.38
1.94
6300
9604
9604
8260
0.236
0.519
0.535
0.197
Se 5 C1C2 (q 2 q) a
C1 5 1 2 0.5a
Iz
Es
Iz
Dz (m3> kN)
Es
1.87 105
3.35 105
7.68 105
4.62 105
17.52 105
Dz
q
21
b 5 1 2 0.5a
b 5 0.915
q2q
145 2 21
Assume the time for creep is 10 years. So,
C2 5 1 1 0.2loga
10
b 5 1.4
0.1
Hence,
Se (0.915)(1.4)(145 21)(17.52 105) 2783 105 m 27.83 mm
5.13
■
Settlement of Foundation on Sand Based
on Standard Penetration Resistance
Meyerhof’s Method
Meyerhof (1956) proposed a correlation for the net bearing pressure for foundations with
the standard penetration resistance, N60. The net pressure has been defined as
qnet 5 q 2 gDf
where q 5 stress at the level of the foundation.
According to Meyerhof’s theory, for 25 mm (1 in.) of estimated maximum settlement,
N60
(for B # 1.22 m)
0.08
(5.57)
N60 B 1 0.3 2
¢
≤ (for B . 1.22 m)
0.125
B
(5.58)
qnet (kN>m2 ) 5
and
qnet (kN>m2 ) 5
264 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Since the time that Meyerhof proposed his original correlations, researchers have
observed that its results are rather conservative. Later, Meyerhof (1965) suggested that
the net allowable bearing pressure should be increased by about 50%. Bowles (1977)
proposed that the modified form of the bearing equations be expressed as
qnet (kN>m2 ) 5
N60
Se
Fa b
2.5 d 25
(for B # 1.22 m)
(5.59)
and
qnet (kN>m2 ) 5
N60 B 1 0.3 2
Se
¢
≤ Fd a b (for B . 1.22 m)
0.08
B
25
(5.60)
where
Fd 5 depth factor 5 1 1 0.33(Df > B)
B 5 foundation width, in meters
Se 5 settlement, in mm
Hence,
Se (mm) 5
1.25qnet (kN>m2 )
N60Fd
(for B < 1.22 m)
(5.61)
and
Se (mm) 5
2
2qnet (kN>m2 )
B
¢
≤
N60Fd
B 1 0.3
(for B . 1.22 m)
(5.62)
The N60 referred to in the preceding equations is the standard penetration resistance
between the bottom of the foundation and 2B below the bottom.
Burland and Burbidge’s Method
Burland and Burbidge (1985) proposed a method of calculating the elastic settlement of
sandy soil using the field standard penetration number, N60 . (See Chapter 2.) The method
can be summarized as follows:
1. Variation of Standard Penetration Number with Depth
Obtain the field penetration numbers (N60 ) with depth at the location of the
foundation. The following adjustments of N60 may be necessary, depending on
the field conditions:
For gravel or sandy gravel,
N60(a) < 1.25 N60
(5.63)
For fine sand or silty sand below the groundwater table and N60 . 15,
N60(a) < 15 1 0.5(N60 2 15)
(5.64)
where N60(a) 5 adjusted N60 value.
2. Determination of Depth of Stress Influence (zzr)
In determining the depth of stress influence, the following three cases may arise:
265
5.13 Settlement of Foundation on Sand Based on Standard Penetration Resistance
Case I. If N60 [or N60(a)] is approximately constant with depth, calculate zr from
zr
B 0.75
5 1.4¢ ≤
BR
BR
(5.65)
where
BR 5 reference width 5 0.3 m (if B is in m)
B 5 width of the actual foundation
Case II. If N60 [or N60(a)] is increasing with depth, use Eq. (5.65) to calculate zr.
Case III. If N60 [or N60(a)] is decreasing with depth, zr 5 2B or to the bottom of soft soil
layer measured from the bottom of the foundation (whichever is smaller).
3. Calculation of Elastic Settlement Se
The elastic settlement of the foundation, Se , can be calculated from
2
Se
5 a1a2a3 E
BR
L
1.25¢ ≤
B
L
0.25 1 ¢ ≤
B
U ¢
B 0.7 qr
≤ ¢ ≤
pa
BR
(5.66)
where
a1 5 a constant
a2 5 compressibility index
a3 5 correction for the depth of influence
pa 5 atmospheric pressure 5 100 kN>m2
L 5 length of the foundation
Table 5.12 summarizes the values of q , 1, 2, and 3 to be used in Eq. (5.70) for
2
2
various types of soils. Note that, in this table, N
60 or N 60(a) average value of N60 or N60(a) in
the depth of stress influence.
Table 5.12 Summary of
1,
2,
and
3
Soil type
q
␣1
␣2
Normally consolidated
sand
qnet
0.14
1.71
2 or N
2 4 1.4
3N
60
60(a)
a3 5
Overconsolidated
sand (qnet c )
qnet
0.047
0.57
2 or N
2 4 1.4
3N
60
60(a)
or
H
H
a2 2 9 b
zr
z
(if H z )
3
1 (if H
z)
where H depth of
compressible layer
where
c preconsolidation
pressure
Overconsolidated
sand (qnet c )
␣3
qnet 0.67c
0.14
0.57
2 or N
2 4 1.4
3N
60
60(a)
266 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Example 5.8
A shallow foundation measuring 1.75 m 3 1.75 m is to be constructed over a layer
of sand. Given Df 5 1 m; N60 is generally increasing with depth; N60 in the depth of
stress influence 5 10, qnet 5 120 kN> m2. The sand is normally consolidated. Estimate the elastic settlement of the foundation. Use the Burland and Burbidge
method.
Solution
From Eq. (5.69),
zr
B 0.75
5 1.4a b
BR
BR
Depth of stress influence,
zr 5 1.4a
B 0.75
1.75 0.75
b BR 5 (1.4) (0.3) a
b
< 1.58 m
BR
0.3
From Eq. (5.70),
2
L
1.25a b
Se
B
B 0.7 qr
5 a1a2a3 D
T a b a b
pa
BR
L
BR
0.25 1 a b
B
For normally consolidated sand (Table 5.12),
a1 5 0.14
1.71
1.71
a2 5
5
5 0.068
1.4
(N60 )
(10) 1.4
a3 5 1
qr 5 qnet 5 120 kN>m2
So,
(1.25) a
2
1.75
b
Se
1.75
1.75 0.7 120
5 (0.14) (0.068) (1) D
b a
b
T a
0.3
0.3
100
1.75
0.25 1 a
b
1.75
Se < 0.0118 m 5 11.8 mm
■
5.14 Settlement in Granular Soil Based on Pressuremeter Test (PMT)
267
Example 5.9
Solve Example 5.8 using Meyerhof ’s method.
Solution
From Eq. (5.66),
2
2qnet
B
¢
≤
(N60 ) (Fd ) B 1 0.3
Fd 5 1 1 0.33(Df>B) 5 1 1 0.33(1>1.75) 5 1.19
Se 5
Se 5
5.14
2
(2) (120)
1.75
¢
≤ 5 14.7 mm
(10) (1.19) 1.75 1 0.3
■
Settlement in Granular Soil Based
on Pressuremeter Test (PMT)
Briaud (2007) proposed a method based on Pressuremeter tests (Section 2.22) from which
the load-settlement diagrams of foundations can be derived. The following is a step-by-step
procedure for performing the analysis.
Step 1.
Step 2.
Step 3.
Conduct Pressuremeter tests at varying depths at the desired location and
obtain plots of pp (pressure in the measuring cell for cavity expansion;
see Figure 2.32) versus R/Ro (Ro initial radius of the PMT cavity, and
R increase in the cavity radius), as shown in Figure 5.24a.
Extend the straight line part of the PMT curve to zero pressure and
shift the vertical axis, as shown in Figure 5.24a. Re-zero the R/Ro axis.
Draw a strain influence factor diagram for the desired foundation (Section
5.12). Using all Pressuremeter test curves within the depth of influence,
develop a mean PMT curve. Referring to Figure 5.24b, this can be done as
follows:
For each value of R/Ro, let the pp values be pp(1), pp(2), pp(3), . . . . The
mean value of pp can be obtained as
pp(m) 5
A1
A2
A3
pp(1) 1
pp(2) 1
p
1 c#
A
A
A p(3)
(5.67)
where
A1, A2, A3 areas tributary to each test under the strain influence factor
diagram
A A1 A2 A3 c#
(5.68)
Step 4.
Based on the results of Step 3, develop a mean pp(m) versus R/Ro plot
(Figure 5.24c).
268 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
pp
(a)
New origin for R
R
Ro
Ro
Strain influence factor, Iz
A1
A2
(b)
A3
PMT
1
2
3
Depth, z
pp(m)
(c)
R
Ro
Figure 5.24 (a) Plot of pp versus R/Ro; (b) averaging the pressuremeter curves
within the foundation zone of influence; (c) plot of pp(m) versus R/Ro
Step 5.
The mean PMT curve now can be used to develop the load-settlement plot
for the foundation via the following equations.
Se
DR
5 0.24
B
Ro
(5.69)
qo 5 fL>Bfefdfb,dGpp(m)
(5.70)
and
where
Se 5 elastic settlement of the foundation
B 5 width of foundation
L 5 length of foundation
qo 5 net load per unit area on the foundation
G 5 gamma function linking qo and pp(m)
269
5.14 Settlement in Granular Soil Based on Pressuremeter Test (PMT)
B
fL>B 5 shape factor 5 0.8 1 0.2a b
L
(5.71)
e
fe 5 eccentricity factor 5 1 2 0.33a b (center)
B
0.5
e
fe 5 eccentricity factor 5 1 2 a b (edge)
B
d(deg) 2
fd 5 load inclination factor 5 1 2 c
d (center)
90
d(deg) 0.5
fd 5 load inclination factor 5 1 2 c
d (edge)
360
d 0.1
fb,d 5 slope factor 5 0.8a1 1 b (3H:1V slope)
B
d 0.15
fb,d 5 slope factor 5 0.7a1 1 b
( 2H:1V slope)
B
(5.72)
(5.73)
(5.74)
(5.75)
(5.76)
(5.77)
inclination of load with respect to the vertical
inclination of a slope with the horizontal if the foundation is located
on top of a slope
d distance of the edge of the foundation from the edge of the slope
The parameters , , d, and e are defined in Figure 5.25. Figure 5.26 shows the design plot
DR
for with Se/B or 0.24
.
Ro
Γ
0
0
1
2
0.02
ΔR
4.2 Ro
or
Se
B
0.04
0.06
Q
e
δ
β
d
0.08
Foundation
B×L
B
Figure 5.25 Definition of parameters—,
L, d, , , e, and 0.1
Figure 5.26 Variation of with
Se/B 0.24 R/Ro
3
270 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Step 6.
Based on the values of B/L, e/B, , and d/B, calculate the values of fL/B, fe,
f, and f,d as needed. Let
f 5 (fL>B ) (fe ) (fd ) (fb,d )
(5.78)
qo 5 fGpp(m)
(5.79)
Thus,
Step 7.
Step 8.
Now prepare a table, as shown in Table 5.13.
Complete Table 5.13 as follows:
a. Column 1—Assume several values of R/Ro.
b. Column 2—For given values of R/Ro, obtain pp(m) from Figure 5.24c.
c. Column 3—From Eq. (5.73), calculate the values of Se/B from values of
R/Ro given in Column 1.
d. Column 4—With known values of B, calculate the values of Se.
e. Column 5—From Figure 5.26, obtain the desired values of .
f. Column 6—Use Eq. (5.83) to obtain qo.
g. Now plot a graph of Se (Column 4) versus qo (Column 6) from which
the magnitude of Se for a given qo can be determined.
Table 5.13 Calculations to Obtain the Load-Settlement Plot
R/Ro
(1)
pp(m)
(2)
Se/B
(3)
Se
(4)
(5)
qo
(6)
Example 5.10
A spread footing, shown in Figure 5.27a with a width of 4 m and a length of 20 m,
serves as a bridge abutment foundation. The soil is medium dense sand. A 16,000 kN
vertical load acts on the footing. The active pressure on the abutment wall develops a
1,600 kN horizontal load. The resultant reaction force due to the vertical and horizontal
load is applied at an eccentricity of 0.13 m. PMT testing at the site produced a mean
Pressuremeter curve characterizing the soil and is shown in Figure 5.27b. What is the
settlement at the current loading?
5.14 Settlement in Granular Soil Based on Pressuremeter Test (PMT)
V = 16,000 kN
4m
1
H = 1600 kN
3
d=3 m
L = 20 m
e = 0.13 m
B=4 m
(a)
1800
1600
pp (m) (kN/m2)
1400
pp (m)
1200
1000
800
600
400
200
ΔR/Ro
(kN/m2)
0.002
0.005
0.01
0.02
0.04
0.07
0.1
0.2
50
150
250
450
800
1200
1400
1700
0
0
0.05
0.1
ΔR/Ro
(b)
0.15
0.2
Figure 5.27
Solution
Given: B 4 m, L 20 m, d 3 m, and slope 3H:1V.
So
4
B
fL>B 5 0.8 1 0.2a b 5 0.8 1 0.2a b 5 0.84
L
20
e
0.13
fe(center) 5 1 2 0.33a b 5 1 2 0.33a
b 5 0.99
B
4
fd(center) 5 1 2 a
d 2
b
90
271
272 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
d 5 tan21 a
fd 5 1 2 a
H
1600
b 5 tan21 a
b 5 5.71°
V
16,000
5.71 2
b 5 0.996
90
fb,d 5 0.8a1 1
d 0.1
3 0.1
b 5 0.8a1 1 b 5 0.846
B
4
f fL/Bfe f f,d (0.84)(0.99)(0.996)(0.845) 0.7
Now the following table can be prepared.
R /Ro
(1)
pp(m)
(kN/m2)
(2)
Se/B
(3)
Se (mm)
(4)
(5)
qo (kN/m2)
(6)
Qo (MN)
(7)
0.002
0.005
0.01
0.02
0.04
0.07
0.10
0.20
50
150
250
450
800
1200
1400
1700
0.0005
0.0012
0.0024
0.0048
0.0096
0.0168
0.024
0.048
2.0
4.8
9.6
19.2
38.4
67.2
96.0
192.0
2.27
2.17
2.07
1.83
1.40
1.17
1.07
0.90
79.45
227.85
362.25
576.45
784.00
982.8
1048.6
1071.0
6.36
18.23
28.98
46.12
62.72
78.62
83.89
85.68
Note: Columns 1 and 2: From Figure 5.27b
Column 3: (Column 1)(0.24) Se/B
Column 4: (Column 3)(B 4000 mm) Se
Column 5: From Figure 5.26
Column 6: f pp(m) (0.7)()pp(m) qo
Column 7: (Column 6)(B L) Qo
Figure 5.28 shows the plot of Qo versus Se. From this plot it can be seen that, for a
vertical loading of 16,000 kN (16 MN), the value of Se ⬇ 4.2 mm.
Qo (MN)
0
0
20
40
60
80
100
20
40
Se (mm)
60
80
100
120
140
160
180
200
Figure 5.28
■
68125_05_ch5_p223-290.qxd
3/12/10
11:16 AM
Page 273
5.15 Primary Consolidation Settlement Relationships
273
Consolidation Settlement
5.15
Primary Consolidation Settlement Relationships
As mentioned before, consolidation settlement occurs over time in saturated clayey
soils subjected to an increased load caused by construction of the foundation. (See
Figure 5.29.) On the basis of the one-dimensional consolidation settlement equations
given in Chapter 1, we write
Sc(p) 5 3ezdz
(5.80)
where
ez 5 vertical strain
De
5
1 1 eo
De 5 change of void ratio
5 f(sor , scr , and Dsr)
So,
Sc(p) 5
CcHc
sor 1 Dsav
r
log
1 1 eo
sor
(for normally consolidated
clays)
qo
Stress
increase,
Groundwater table
t
Clay layer Hc
m
b
Depth, z
Figure 5.29 Consolidation settlement calculation
(5.81)
274 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Sc(p) 5
CsHc
sor 1 Dsav
r
log
1 1 eo
sor
(for overconsolidated clays
with sor 1 Dsav
r , scr )
(5.82)
Sc(p) 5
CsHc
scr
CcHc
sor 1 Dsav
r
log
1
log
1 1 eo
sor
1 1 eo
scr
(for overconsolidated clays
with sor , scr , sor 1 Dsav
r)
(5.83)
where
sor 5 average effective pressure on the clay layer before the construction of the foundation
Dsav
r 5 average increase in effective pressure on the clay layer caused by the
construction of the foundation
scr 5 preconsolidation pressure
eo 5 initial void ratio of the clay layer
Cc 5 compression index
Cs 5 swelling index
Hc 5 thickness of the clay layer
The procedures for determining the compression and swelling indexes were discussed in
Chapter 1.
Note that the increase in effective pressure, Dsr, on the clay layer is not constant with
depth: The magnitude of Dsr will decrease with the increase in depth measured from the bottom of the foundation. However, the average increase in pressure may be approximated by
Dsav
r 5 16 ( Dstr 1 4Dsm
r 1 Dsbr )
(5.84)
where Dstr , Dsm
r , and Dsbr are, respectively, the effective pressure increases at the top,
middle, and bottom of the clay layer that are caused by the construction of the foundation.
The method of determining the pressure increase caused by various types of foundation
load using Boussinesq’s solution is discussed in Sections 5.2 through 5.6. Dsav
r can also be
directly obtained from the method presented in Section 5.5.
5.16
Three-Dimensional Effect on Primary
Consolidation Settlement
The consolidation settlement calculation presented in the preceding section is based on
Eqs. (1.61), (1.63), and (1.65). These equations, as shown in Chapter 1, are in turn based
on one-dimensional laboratory consolidation tests. The underlying assumption is that the
increase in pore water pressure, Du, immediately after application of the load equals the
increase in stress, Ds, at any depth. In this case,
De
Sc(p) 2oed 5 3
dz 5 3mv Ds(1)
r dz
1 1 eo
where
Sc(p) 2oed 5 consolidation settlement calculated by using Eqs. (1.61), (1.63), and (1.65)
Ds(1)
r 5 effective vertical stress increase
mv 5 volume coefficient of compressibility (see Chapter 1)
5.16 Three-Dimensional Effect on Primary Consolidation Settlement
275
Flexible
circular load
Clay
(1)
Hc
(3)
z
(3)
Figure 5.30 Circular foundation on a clay layer
In the field, however, when a load is applied over a limited area on the ground surface,
such an assumption will not be correct. Consider the case of a circular foundation on a clay
layer, as shown in Figure 5.30. The vertical and the horizontal stress increases at a point in
the layer immediately below the center of the foundation are Ds(1) and Ds(3) , respectively.
For a saturated clay, the pore water pressure increase at that depth (see Chapter 1) is
Du 5 Ds(3) 1 A3Ds(1) 2 Ds(3) 4
(5.85)
Sc(p) 5 3mv Du dz 5 3 (mv ) 5Ds(3) 1 A3Ds(1) 2 Ds(3) 46 dz
(5.86)
where A 5 pore water pressure parameter. For this case,
Thus, we can write
Hc
Kcir 5
Sc(p)
Sc(p) 2oed
Hc
3 mv Du dz
5
0
Hc
r dz
3 Ds(3)
5 A 1 (1 2 A) D
r dz
3 mv Ds(1)
0
Hc
T
(5.87)
r dz
3 Ds(1)
0
0
where Kcir 5 settlement ratio for circular foundations.
The settlement ratio for a continuous foundation, Kstr , can be determined in a manner similar to that for a circular foundation. The variation of Kcir and Kstr with A and Hc>B
is given in Figure 5.31. (Note: B 5 diameter of a circular foundation, and B 5 width of a
continuous foundation.)
The preceding technique is generally referred to as the Skempton–Bjerrum modification (1957) for a consolidation settlement calculation.
Leonards (1976) examined the correction factor Kcr for a three-dimensional consolidation effect in the field for a circular foundation located over overconsolidated clay.
Referring to Figure 5.30, we have
Sc(p) 5 Kcr(OC) Sc(p) 2oed
(5.88)
B
≤
Hc
(5.89)
where
Kcr(OC) 5 f¢OCR,
276 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
1.0
H c/B =
Settlement ratio
0.8
0.25
0.25 0.5
0.5
0.6
1.0
2.0
1.0
0.4
2.0
Circular
foundation
0.2
Continuous
foundation
0
0
0.2
0.4
0.6
0.8
Pore water pressure parameter, A
1.0
Figure 5.31 Settlement
ratios for circular (Kcir ) and
continuous (Kstr )
foundations
in which
OCR 5 overconsolidation ratio 5
scr
sor
(5.90)
where
scr 5 preconsolidation pressure
sor 5 present average effective pressure
The interpolated values of Kcr(OC) from Leonard’s 1976 work are given in Table 5.14.
Table 5.14 Variation of Kcr(OC) with OCR and B>Hc
Kcr (OC )
OCR
B/Hc ⴝ 4.0
B/Hc ⴝ 1.0
B/Hc ⴝ 0.2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
1
0.986
0.972
0.964
0.950
0.943
0.929
0.914
0.900
0.886
0.871
0.864
0.857
0.850
0.843
0.843
1
0.957
0.914
0.871
0.829
0.800
0.757
0.729
0.700
0.671
0.643
0.629
0.614
0.607
0.600
0.600
1
0.929
0.842
0.771
0.707
0.643
0.586
0.529
0.493
0.457
0.429
0.414
0.400
0.386
0.371
0.357
5.16 Three-Dimensional Effect on Primary Consolidation Settlement
277
Example 5.11
A plan of a foundation 1 m 3 2 m is shown in Figure 5.32. Estimate the consolidation
settlement of the foundation, taking into account the three-dimensional effect. Given:
A 5 0.6.
Solution
The clay is normally consolidated. Thus,
Sc(p)2oed 5
CcHc
sor 1 Dsav
r
log
1 1 eo
sor
so
sor 5 (2.5) (16.5) 1 (0.5) (17.5 2 9.81) 1 (1.25) (16 2 9.81)
5 41.25 1 3.85 1 7.74 5 52.84 kN>m2
From Eq. (5.84),
Dsav
r 5 16 (Dstr 1 4Dsm
r 1 Dsbr )
Now the following table can be prepared (Note: L 5 2 m; B 5 1 m):
a
b
m1 ⴝ L/B
z(m)
z/(B/2) ⴝ n1
Ica
⌬s9 ⴝ qoIcb
2
2
2
2
2 1 2.5>2 5 3.25
2 1 2.5 5 4.5
4
6.5
9
0.190
< 0.085
0.045
28.5 5 Dstr
r
12.75 5 Dsm
6.75 5 Dsbr
Table 5.3
Eq. (5.10)
q0 150 kN/m2 (net stress increase)
1m
BLlm2m
1.5 m
Sand
16.5 kN/m3
Groundwater table
0.5 m
2.5 m
Sand
sat 17.5 kN/m3
Normally consolidated clay
ea 0.8
16 kN/m3
s 6,000 kN/m2 Ce 0.32
s 0.5
Cs 0.09
Figure 5.32 Calculation of primary consolidation settlement for a
foundation
278 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Now,
Dsav
r 5 16 (28.5 1 4 3 12.75 1 6.75) 5 14.38 kN>m2
so
Sc(p)2oed 5
(0.32) (2.5)
52.84 1 14.38
log ¢
≤ 5 0.0465 m
1 1 0.8
52.84
5 46.5 mm
Now assuming that the 2:1 method of stress increase (see Figure 5.5) holds good, the
area of distribution of stress at the top of the clay layer will have dimensions
B9 5 width 5 B 1 z 5 1 1 (1.5 1 0.5) 5 3 m
and
L9 5 width 5 L 1 z 5 2 1 (1.5 1 0.5) 5 4 m
The diameter of an equivalent circular area, Beq, can be given as
p 2
Beq 5 BrLr
4
so that
Beq 5
Å
4BrLr
(4) (3) (4)
5
5 3.91 m
p
p
Å
Also,
Hc
2.5
5
5 0.64
Beq
3.91
From Figure 5.31, for A 5 0.6 and Hc>Beq 5 0.64, the magnitude of Kcr < 0.78.
Hence,
Se(p) 5 KcrSe(p) – oed 5 (0.78)(46.5) < 36.3 mm
5.17
■
Settlement Due to Secondary Consolidation
At the end of primary consolidation (i.e., after the complete dissipation of excess pore
water pressure) some settlement is observed that is due to the plastic adjustment of soil
fabrics. This stage of consolidation is called secondary consolidation. A plot of deformation against the logarithm of time during secondary consolidation is practically linear
Void ratio, e
5.17 Settlement Due to Secondary Consolidation
C 279
e
t
log t2
1
ep
e
t1
Time, t (log scale)
t2
Figure 5.33 Variation of e with
log t under a given load increment,
and definition of secondary compression index
as shown in Figure 5.33. From the figure, the secondary compression index can be
defined as
Ca 5
De
De
5
log t2 2 log t1
log (t2>t1 )
(5.91)
where
Ca 5 secondary compression index
De 5 change of void ratio
t1 , t2 5 time
The magnitude of the secondary consolidation can be calculated as
Sc(s) 5 C ar Hc log(t2>t1 )
(5.92)
where
C ar 5 Ca> (1 1 ep )
ep 5 void ratio at the end of primary consolidation
Hc 5 thickness of clay layer
(5.93)
Mesri (1973) correlated C ar with the natural moisture content (w) of several soils,
from which it appears that
C ar < 0.0001w
(5.94)
where w 5 natural moisture content, in percent. For most overconsolidated soils, C ar varies
between 0.0005 to 0.001.
Mesri and Godlewski (1977) compiled the magnitude of Ca> Cc (Cc 5 compression
index) for a number of soils. Based on their compilation, it can be summarized that
•
For inorganic clays and silts:
Ca>Cc < 0.04 6 0.01
280 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
•
For organic clays and silts:
Ca>Cc < 0.05 6 0.01
•
For peats:
Ca>Cc < 0.075 6 0.01
Secondary consolidation settlement is more important in the case of all organic and
highly compressible inorganic soils. In overconsolidated inorganic clays, the secondary
compression index is very small and of less practical significance.
There are several factors that might affect the magnitude of secondary consolidation,
some of which are not yet very clearly understood (Mesri, 1973). The ratio of secondary
to primary compression for a given thickness of soil layer is dependent on the ratio of the
stress increment, Dsr, to the initial effective overburden stress, sor . For small Dsr>sor
ratios, the secondary-to-primary compression ratio is larger.
5.18
Field Load Test
The ultimate load-bearing capacity of a foundation, as well as the allowable bearing
capacity based on tolerable settlement considerations, can be effectively determined
from the field load test, generally referred to as the plate load test. The plates that are
used for tests in the field are usually made of steel and are 25 mm thick and 150 mm
to 762 mm in diameter. Occasionally, square plates that are 305 mm 3 305 mm are
also used.
To conduct a plate load test, a hole is excavated with a minimum diameter of 4B (B
is the diameter of the test plate) to a depth of Df , the depth of the proposed foundation. The
plate is placed at the center of the hole, and a load that is about one-fourth to one-fifth of the
estimated ultimate load is applied to the plate in steps by means of a jack. A schematic diagram
of the test arrangement is shown in Figure 5.34a. During each step of the application of the
load, the settlement of the plate is observed on dial gauges. At least one hour is allowed to
elapse between each application. The test should be conducted until failure, or at least
until the plate has gone through 25 mm (1 in.) of settlement. Figure 5.34b shows the
nature of the load–settlement curve obtained from such tests, from which the ultimate
load per unit area can be determined. Figure 5.35 shows a plate load test conducted in
the field.
For tests in clay,
qu(F) 5 qu(P)
(5.95)
where
qu(F) 5 ultimate bearing capacity of the proposed foundation
qu(P) 5 ultimate bearing capacity of the test plate
Equation (5.95) implies that the ultimate bearing capacity in clay is virtually independent
of the size of the plate.
5.18 Field Load Test
281
Reaction
beam
Jack
Dial
gauge
Test plate
diameter
B
Anchor
pile
At least
4B
(a)
Load/unit area
Settlement
Figure 5.34 Plate load test: (a) test
arrangement; (b) nature of load–
settlement curve
(b)
For tests in sandy soils,
qu(F) 5 qu(P)
BF
BP
(5.96)
where
BF 5 width of the foundation
BP 5 width of the test plate
The allowable bearing capacity of a foundation, based on settlement considerations
and for a given intensity of load, qo , is
SF 5 SP
BF
BP
(for clayey soil)
(5.97)
and
SF 5 SP ¢
2
2BF
≤
BF 1 BP
(for sandy soil)
(5.98)
282 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Figure 5.35 Plate load test in the field (Courtesy of Braja M. Das, Henderson, NV)
5.19
Presumptive Bearing Capacity
Several building codes (e.g., the Uniform Building Code, Chicago Building Code, and New
York City Building Code) specify the allowable bearing capacity of foundations on various
types of soil. For minor construction, they often provide fairly acceptable guidelines.
However, these bearing capacity values are based primarily on the visual classification of
near-surface soils and generally do not take into consideration factors such as the stress history of the soil, the location of the water table, the depth of the foundation, and the tolerable
settlement. So, for large construction projects, the codes’ presumptive values should be used
only as guides.
5.20
Tolerable Settlement of Buildings
In most instances of construction, the subsoil is not homogeneous and the load carried
by various shallow foundations of a given structure can vary widely. As a result, it is reasonable to expect varying degrees of settlement in different parts of a given building.
5.20 Tolerable Settlement of Buildings
283
L
lAB
B
C
D
A
A
E
E
max
ST (max)
B
D
ST (max)
C
max
Figure 5.36 Definition of parameters
for differential settlement
The differential settlement of the parts of a building can lead to damage of the superstructure. Hence, it is important to define certain parameters that quantify differential
settlement and to develop limiting values for those parameters in order that the resulting
structures be safe. Burland and Worth (1970) summarized the important parameters
relating to differential settlement.
Figure 5.36 shows a structure in which various foundations, at A, B, C, D, and
E, have gone through some settlement. The settlement at A is AAr, at B is BBr, etc.
Based on this figure, the definitions of the various parameters are as follows:
ST 5 total settlement of a given point
DST 5 difference in total settlement between any two points
a 5 gradient between two successive points
DST(ij)
b 5 angular distortion 5
lij
(Note: lij 5 distance between points i and j)
v 5 tilt
D 5 relative deflection (i.e. movement from a straight line joining two reference
points)
D
5 deflection ratio
L
Since the 1950s, various researchers and building codes have recommended allowable values for the preceding parameters. A summary of several of these recommendations
is presented next.
In 1956, Skempton and McDonald proposed the following limiting values for
maximum settlement and maximum angular distortion, to be used for building purposes:
Maximum settlement, ST(max)
In sand
In clay
32 mm
45 mm
284 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
Maximum differential settlement, DST(max)
Isolated foundations in sand
Isolated foundations in clay
Raft in sand
Raft in clay
Maximum angular distortion, bmax
51 mm
76 mm
51–76 mm
76–127 mm
1>300
On the basis of experience, Polshin and Tokar (1957) suggested the following allowable
deflection ratios for buildings as a function of L>H, the ratio of the length to the height of
a building:
D>L 5 0.0003 for L>H # 2
D>L 5 0.001 for L>H 5 8
The 1955 Soviet Code of Practice gives the following allowable values:
Type of building
L ,H
Multistory buildings and
civil dwellings
<3
0.0003 (for sand)
0.0004 (for clay)
>5
0.0005 (for sand)
0.0007 (for clay)
One-story mills
D ,L
0.001 (for sand and clay)
Bjerrum (1963) recommended the following limiting angular distortion, bmax for
various structures:
Category of potential damage
Safe limit for flexible brick wall (L>H . 4)
Danger of structural damage to most buildings
Cracking of panel and brick walls
Visible tilting of high rigid buildings
First cracking of panel walls
Safe limit for no cracking of building
Danger to frames with diagonals
bmax
1>150
1>150
1>150
1>250
1>300
1>500
1>600
If the maximum allowable values of bmax are known, the magnitude of the allowable
ST(max) can be calculated with the use of the foregoing correlations.
The European Committee for Standardization has also provided limiting values for serviceability and the maximum accepted foundation movements. (See
Table 5.15.)
Problems
285
Table 5.15 Recommendations of European Committee for Standardization on Differential
Settlement Parameters
Item
Limiting values for
serviceability
(European Committee
for Standardization,
1994a)
Maximum acceptable
foundation movement
(European Committee
for Standardization, 1994b)
Parameter
ST
DST
b
ST
DST
b
Magnitude
Comments
25 mm
50 mm
5 mm
10 mm
20 mm
1>500
50
20
<1>500
Isolated shallow foundation
Raft foundation
Frames with rigid cladding
Frames with flexible cladding
Open frames
—
Isolated shallow foundation
Isolated shallow foundation
—
Problems
A flexible circular area is subjected to a uniformly distributed load of 150 kN> m2.
The diameter of the loaded area is 2 m. Determine the stress increase in a soil mass
at a point located 3 m below the center of the loaded area.
a. by using Eq. (5.3)
b. by using Eq. (5.28). Use s 5 0
5.2 Refer to Figure 5.4, which shows a flexible rectangular area. Given: B1 5 1.22 m,
B2 5 1.83 m, L1 5 2.44 m, and L2 5 3.05 m. If the area is subjected to a uniform
load of 3000 lb> ft2, determine the stress increase at a depth of 3.05 m located
immediately below point O.
5.3 Repeat Problem 5.2 with the following: B1 5 1.22 m, B2 5 3.05 m, L1 5 2.44 m, L2 5
3.66 m, and the uniform load on the flexible area 5 120 kN/m2. Determine the stress
increase below point O at a depth of 6.1 m. Use Eq. (5.29) and s 5 0.
5.4 Using Eq. (5.10), determine the stress increase (Ds) at z 5 3.05 m below the center of the area described in Problem 5.2.
5.5 Refer to Figure P5.5. Using the procedure outlined in Section 5.5, determine the
average stress increase in the clay layer below the center of the foundation due to the
net foundation load of 900 kN.
5.6 Solve Problem 5.5 using the 2:1 method [Eqs. (5.14) and (5.88)].
5.7 Figure P5.7 shows an embankment load on a silty clay soil layer. Determine the
stress increase at points A, B, and C, which are located at a depth of 5 m below the
ground surface.
5.8 A planned flexible load area (see Figure P5.8) is to be 2 m 3 3.2 m and carries a uniformly distributed load of 210 kN> m2. Estimate the elastic settlement below the center
of the loaded area. Assume that Df 5 1.6 m and H 5 `. Use Eq. (5.33).
5.9 Redo Problem 5.8 assuming that Df 5 1.2 m and H 5 4 m.
5.10 Consider a flexible foundation measuring 1.52 m 3.05 m in a plan on a soft
saturated clay (s 0.5). The depth of the foundation is 1.22 m below the ground
5.1
286 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
900 kN (net load)
Sand
15.7 kN/m3
1.52 m
1.83 m 1.83 m
1.22 m
Groundwater table
Sand
sat 19.24 kN/m3
3.05 m
sat 19.24 kN/m3
eo 0.8
Cc 0.25
Cs 0.06
Preconsolidation pressure 100 kN/m2
Figure P5.5
6m
Center line
1V:2H
1V:2H
10 m
γ = 17 kN/m3
5m
C
B
Df
A
Figure P5.7
210 kN/m2
2 m 3.2 m
Silty sand
Es 8500 kN/m2
s 0.3
H
Rock
Figure P5.8
surface. A rigid rock layer is located at 12.2 m below the bottom of the foundation.
Given: qo 144 kN/m2 and, for the clay, Es 12,938 kN/m2. Determine the average
elastic settlement of the foundation. Use Eq. (5.30).
5.11 Figure 5.16 shows a foundation of 3.05 m 3 1.91 m resting on a sand deposit. The
net load per unit area at the level of the foundation, qo, is 144 kN> m2. For the sand,
ms 5 0.3, Es 5 22,080 kN> m2, Df 5 0.76 m and H 5 9.76 m. Assume that the foundation is rigid, and determine the elastic settlement the foundation would undergo.
Use Eqs. (5.33) and (5.41).
Problems
287
5.12 Repeat Problems 5.11 for a foundation of size 5 1.8 m 3 1.8 m and with
qo 5 190 kN> m2, Df 5 1.0 m, H 5 15 m; and soil conditions of ms 5 0.4, Es 5
15,400 kN> m2, and g 5 17 kN> m3.
5.13 For a shallow foundation supported by a silty clay, as shown in Figure 5.17, the
following are given:
Length, L 5 2 m
Width, B 5 1 m
Depth of foundation, Df 5 1 m
Thickness of foundation, t 5 0.23 m
Load per unit area, qo 5 190 kN> m2
Ef 5 15 3 106 kN> m2
The silty clay soil has the following properties:
H52m
ms 5 0.4
Eo 5 9000 kN> m2
k 5 500 kN> m2> m
Using Eq. (5.46), estimate the elastic settlement of the foundation.
5.14 A plan calls for a square foundation measuring 3 m 3 3 m, supported by a layer
of sand. (See Figure 5.17.) Led Df 5 1.5 m, t 5 0.25 m, Eo 5 16,000 kN> m2, k 5
400 kN> m2> m, ms 5 0.3, H 5 20 m, Ef 5 15 3 106 kN> m2, and qo 5
150 kN> m2. Calculate the elastic settlement. Use Eq. (5.46).
5.15 Solve Problem 5.11 with Eq. (5.49). For the correction factor C2-, use a time of
5 years for creep and, for the unit weight of soil, use g 5 18.08 kN> m3.
5.16 A continuous foundation on a deposit of sand layer is shown in Figure P5.16
along with the variation of the modulus of elasticity of the soil (Es). Assuming g 5
18 kN> m3 and C2 for 10 years, calculate the elastic settlement of the foundation using the strain influence factor method.
1.5 m
q 195 kN/m2
0
Es 6000
2.5 m
Sand
Es(kN/m2)
2
Es 12,000
8
Es 10,000
14
Depth (m)
Figure P5.16
288 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
5.17 Following are the results of standard penetration tests in a granular soil deposit.
Depth
(m)
Standard penetration
number, N60
1.5
3.0
4.5
6.0
7.5
11
10
12
9
14
What will be the net allowable bearing capacity of a foundation planned to be 1.83 m 3
1.83 m? Let Df 5 6.91 m and the allowable settlement 5 25 mm, and use the relationships presented in Eq. (5.60).
5.18 A shallow foundation measuring 1.2 m 3 1.2 m in plan is to be constructed over a
normally consolidated sand layer. Given: Df 5 1 m, N60 increases with depth, N60 (in
the depth of stress influence) 5 8, and qnet 5 210 kN> m2. Estimate the elastic settlement using Burland and Burbidge’s method.
5.19 Refer to Figure 5.25. For a foundation on a layer of sand, given: B 1.52 m, L 3.05 m, d 1.52 m, 26.6°, e 0.152 m, and 109. The Pressuremeter testing
at the site produced a mean Pressuremeter curve for which the pp(m) versus R/Ro
points are as follow:
R/Ro
(1)
pp(m) (lb/in2)
(2)
0.002
0.004
0.008
0.012
0.024
0.05
0.08
0.1
0.2
49.68
166.98
224.94
292.56
475.41
870.09
1225.97
1452.45
2550.24
What should be the magnitude of Qo for a settlement (center) of 25 mm?
5.20 Estimate the consolidation settlement of the clay layer shown in Figure P5.5 using
the results of Problem 5.5.
5.21 Estimate the consolidation settlement of the clay layer shown in Figure P5.5 using
the results of Problem 5.6.
References
AHLVIN, R. G., and ULERY, H. H. (1962). Tabulated Values of Determining the Composite Pattern of
Stresses, Strains, and Deflections beneath a Uniform Load on a Homogeneous Half Space.
Highway Research Board Bulletin 342, pp. 1–13.
BJERRUM, L. (1963). “Allowable Settlement of Structures,” Proceedings, European Conference on
Soil Mechanics and Foundation Engineering, Wiesbaden, Germany, Vol. III, pp. 135 –137.
References
289
BOUSSINESQ, J. (1883). Application des Potentials á L’Étude de L’Équilibre et du Mouvement des
Solides Élastiques, Gauthier-Villars, Paris.
BOWLES, J. E. (1987). “Elastic Foundation Settlement on Sand Deposits,” Journal of Geotechnical
Engineering, ASCE, Vol. 113, No. 8, pp. 846– 860.
BOWLES, J. E. (1977). Foundation Analysis and Design, 2d ed., McGraw-Hill, New York.
BRIAUD, J. L. (2007). “Spread Footings in Sand: Load Settlement Curve Approach,” Journal of
Geotechnical and Geoenvironmental Engineering, American Society of Civil Engineers,
Vol. 133, No. 8, pp. 905–920.
BURLAND, J. B., and BURBIDGE, M. C. (1985). “Settlement of Foundations on Sand and Gravel,”
Proceedings, Institute of Civil Engineers, Part I, Vol. 7, pp. 1325–1381.
CHRISTIAN, J. T., and CARRIER, W. D. (1978). “Janbu, Bjerrum, and Kjaernsli’s Chart Reinterpreted,”
Canadian Geotechnical Journal, Vol. 15, pp. 124–128.
DAS, B. (2008). Advanced Soil Mechanics, 3d ed., Taylor and Francis, London.
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EUROPEAN COMMITTEE FOR STANDARDIZATION (1994a). Basis of Design and Actions on Structures,
Eurocode 1, Brussels, Belgium.
EUROPEAN COMMITTEE FOR STANDARDIZATION (1994b). Geotechnical Design, General Rules—Part 1,
Eurocode 7, Brussels, Belgium.
FOX, E. N. (1948). “The Mean Elastic Settlement of a Uniformly Loaded Area at a Depth below the
Ground Surface,” Proceedings, 2nd International Conference on Soil Mechanics and Foundation Engineering, Rotterdam, Vol. 1, pp. 129–132.
GRIFFITHS, D. V. (1984). “A Chart for Estimating the Average Vertical Stress Increase in an Elastic
Foundation below a Uniformly Loaded Rectangular Area,” Canadian Geotechnical Journal,
Vol. 21, No. 4, pp. 710–713.
JANBU, N., BJERRUM, L., and KJAERNSLI, B. (1956). “Veiledning vedlosning av fundamentering—
soppgaver,” Publication No. 18, Norwegian Geotechnical Institute, pp. 30–32.
LEONARDS, G. A. (1976). Estimating Consolidation Settlement of Shallow Foundations on Overconsolidated Clay, Special Report No. 163, Transportation Research Board, Washington, D.C.,
pp. 13–16.
MAYNE, P. W., and POULOS, H. G. (1999). “Approximate Displacement Influence Factors for Elastic
Shallow Foundations,” Journal of Geotechnical and Geoenvironmental Engineering, ASCE,
Vol. 125, No. 6, pp. 453– 460.
MESRI, G. (1973). “Coefficient of Secondary Compression,” Journal of the Soil Mechanics and
Foundations Division, American Society of Civil Engineers, Vol. 99, No. SM1, pp. 122–137.
MESRI, G., and GODLEWSKI, P. M. (1977). “Time and Stress—Compressibility Interrelationship,”
Journal of Geotechnical Engineering Division, American Society of Civil Engineers,
Vol. 103, No. GT5, pp. 417–430.
MEYERHOF, G. G. (1956). “Penetration Tests and Bearing Capacity of Cohesionless Soils,” Journal of
the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 82,
No. SM1, pp. 1–19.
MEYERHOF, G. G. (1965). “Shallow Foundations,” Journal of the Soil Mechanics and Foundations
Division, American Society of Civil Engineers, Vol. 91, No. SM2, pp. 21–31.
NEWMARK, N. M. (1935). Simplified Computation of Vertical Pressure in Elastic Foundation, Circular 24, University of Illinois Engineering Experiment Station, Urbana, IL.
OSTERBERG, J. O. (1957). “Influence Values for Vertical Stresses in Semi-Infinite Mass Due to
Embankment Loading,” Proceedings, Fourth International Conference on Soil Mechanics and
Foundation Engineering, London, Vol. 1, pp. 393 –396.
POLSHIN, D. E., and TOKAR, R. A. (1957). “Maximum Allowable Nonuniform Settlement of Structures,” Proceedings, Fourth International Conference on Soil Mechanics and Foundation
Engineering, London, Vol. 1, pp. 402–405.
SALGADO, R. (2008). The Engineering of Foundations, McGraw-Hill, New York.
290 Chapter 5: Shallow Foundations: Allowable Bearing Capacity and Settlement
SCHMERTMANN, J. H., HARTMAN, J. P., and BROWN, P. R. (1978). “Improved Strain Influence Factor
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Mechanics of Solids, Dedicated to Stephen Timoshenko, pp. 268–277, MacMillan, New York.
6
6.1
Mat Foundations
Introduction
Under normal conditions, square and rectangular footings such as those described in
Chapters 3 and 4 are economical for supporting columns and walls. However, under certain
circumstances, it may be desirable to construct a footing that supports a line of two or more
columns. These footings are referred to as combined footings. When more than one line of
columns is supported by a concrete slab, it is called a mat foundation. Combined footings
can be classified generally under the following categories:
a. Rectangular combined footing
b. Trapezoidal combined footing
c. Strap footing
Mat foundations are generally used with soil that has a low bearing capacity. A brief
overview of the principles of combined footings is given in Section 6.2, followed by a
more detailed discussion on mat foundations.
6.2
Combined Footings
Rectangular Combined Footing
In several instances, the load to be carried by a column and the soil bearing capacity are
such that the standard spread footing design will require extension of the column foundation beyond the property line. In such a case, two or more columns can be supported on a
single rectangular foundation, as shown in Figure 6.1. If the net allowable soil pressure is
known, the size of the foundation (B 3 L) can be determined in the following manner:
a. Determine the area of the foundation
A5
Q1 1 Q2
qnet(all)
(6.1)
where
Q1 , Q2 5 column loads
qnet(all) 5 net allowable soil bearing capacity
291
292 Chapter 6: Mat Foundations
Q1 Q2
L2
L1
X
L3
Q1
Q2
Section
B qnet(all) /unit length
L
Property
line
B
Plan
Figure 6.1 Rectangular combined footing
b. Determine the location of the resultant of the column loads. From Figure 6.1,
X5
Q2L3
Q1 1 Q2
(6.2)
c. For a uniform distribution of soil pressure under the foundation, the resultant of
the column loads should pass through the centroid of the foundation. Thus,
L 5 2(L2 1 X)
(6.3)
where L 5 length of the foundation.
d. Once the length L is determined, the value of L1 can be obtained as follows:
L1 5 L 2 L2 2 L3
(6.4)
Note that the magnitude of L2 will be known and depends on the location of the
property line.
e. The width of the foundation is then
B5
A
L
(6.5)
Trapezoidal Combined Footing
Trapezoidal combined footing (see Figure 6.2) is sometimes used as an isolated spread
foundation of columns carrying large loads where space is tight. The size of the foundation
that will uniformly distribute pressure on the soil can be obtained in the following manner:
a. If the net allowable soil pressure is known, determine the area of the foundation:
Q1 1 Q2
A5
qnet(all)
6.2 Combined Footings
Q1 Q2
L2
L1
X
L3
Q1
293
Q2
B2 qnet(all) /unit length
B1 qnet(all) /unit length
Section
Property
line
B1
B2
L
Plan
Figure 6.2 Trapezoidal combined footing
From Figure 6.2,
A5
B1 1 B2
L
2
(6.6)
b. Determine the location of the resultant for the column loads:
X5
Q2L3
Q1 1 Q2
c. From the property of a trapezoid,
X 1 L2 5 ¢
B1 1 2B2 L
≤
B1 1 B2 3
(6.7)
With known values of A, L, X, and L2 , solve Eqs. (6.6) and (6.7) to obtain B1 and B2 .
Note that, for a trapezoid,
L
L
, X 1 L2 ,
3
2
294 Chapter 6: Mat Foundations
Section
Section
Strap
Strap
Strap
Strap
Plan
(a)
Plan
(b)
Wall
Section
Strap
Strap
Plan
(c)
Figure 6.3 Cantilever footing—use of strap beam
Cantilever Footing
Cantilever footing construction uses a strap beam to connect an eccentrically loaded column foundation to the foundation of an interior column. (See Figure 6.3). Cantilever footings may be used in place of trapezoidal or rectangular combined footings when
the allowable soil bearing capacity is high and the distances between the columns are large.
6.3
Common Types of Mat Foundations
The mat foundation, which is sometimes referred to as a raft foundation, is a combined
footing that may cover the entire area under a structure supporting several columns and
walls. Mat foundations are sometimes preferred for soils that have low load-bearing capacities, but that will have to support high column or wall loads. Under some conditions,
spread footings would have to cover more than half the building area, and mat foundations
might be more economical. Several types of mat foundations are used currently. Some of
the common ones are shown schematically in Figure 6.4 and include the following:
1. Flat plate (Figure 6.4a). The mat is of uniform thickness.
2. Flat plate thickened under columns (Figure 6.4b).
3. Beams and slab (Figure 6.4c). The beams run both ways, and the columns are
located at the intersection of the beams.
4. Flat plates with pedestals (Figure 6.4d).
5. Slab with basement walls as a part of the mat (Figure 6.4e). The walls act as stiffeners
for the mat.
6.3 Common Types of Mat Foundations
295
Section
Section
Section
Plan
Plan
Plan
(a)
(b)
(c)
Section
Section
Plan
Plan
(e)
(d)
Figure 6.4 Common types of mat foundations
Mats may be supported by piles, which help reduce the settlement of a structure built
over highly compressible soil. Where the water table is high, mats are often placed over
piles to control buoyancy. Figure 6.5 shows the difference between the depth Df and the
width B of isolated foundations and mat foundations. Figure 6.6 shows a flat-plate mat
foundation under construction.
Df
B
Df
B
Figure 6.5 Comparison of isolated foundation and mat
foundation (B 5 width, Df 5 depth)
296 Chapter 6: Mat Foundations
Figure 6.6 A flat plate mat foundation under construction (Courtesy of Dharma Shakya,
Geotechnical Solutions, Inc., Irvine, California)
6.4
Bearing Capacity of Mat Foundations
The gross ultimate bearing capacity of a mat foundation can be determined by the same
equation used for shallow foundations (see Section 3.6), or
qu 5 crNcFcsFcdFci 1 qNqFqsFqdFqi 1 12gBNgFgsFgdFgi
[Eq. (3.19)]
(Chapter 3 gives the proper values of the bearing capacity factors, as well as the shape
depth, and load inclination factors.) The term B in Eq. (3.19) is the smallest dimension of
the mat. The net ultimate capacity of a mat foundation is
qnet(u) 5 qu 2 q
[Eq. (3.14)]
A suitable factor of safety should be used to calculate the net allowable bearing
capacity. For mats on clay, the factor of safety should not be less than 3 under dead load
or maximum live load. However, under the most extreme conditions, the factor of safety
should be at least 1.75 to 2. For mats constructed over sand, a factor of safety of 3 should
normally be used. Under most working conditions, the factor of safety against bearing
capacity failure of mats on sand is very large.
For saturated clays with f 5 0 and a vertical loading condition, Eq. (3.19) gives
qu 5 cuNcFcsFcd 1 q
(6.8)
6.4 Bearing Capacity of Mat Foundations
297
where cu 5 undrained cohesion. (Note: Nc 5 5.14, Nq 5 1, and Ng 5 0.)
From Table 3.4, for f 5 0,
Fcs 5 1 1
B Nq
B
1
0.195B
¢ ≤ 5 1 1 ¢ ≤¢
≤ 511
L Nc
L 5.14
L
and
Fcd 5 1 1 0.4 ¢
Df
B
≤
Substitution of the preceding shape and depth factors into Eq. (6.8) yields
qu 5 5.14cu ¢ 1 1
Df
0.195B
≤ ¢1 1 0.4 ≤ 1 q
L
B
(6.9)
Hence, the net ultimate bearing capacity is
qnet(u) 5 qu 2 q 5 5.14cu ¢ 1 1
Df
0.195B
≤ ¢ 1 1 0.4 ≤
L
B
(6.10)
For FS 5 3, the net allowable soil bearing capacity becomes
qnet(all) 5
qu(net)
FS
5 1.713cu ¢ 1 1
Df
0.195B
≤ ¢ 1 1 0.4 ≤
L
B
(6.11)
The net allowable bearing capacity for mats constructed over granular soil deposits
can be adequately determined from the standard penetration resistance numbers. From
Eq. (5.64), for shallow foundations,
qnet (kN>m2 ) 5
N60 B 1 0.3 2
Se
¢
≤ Fd ¢ ≤
0.08
B
25
[Eq. (5.64)]
where
N60 5 standard penetration resistance
B 5 width (m)
Fd 5 1 1 0.33(Df>B) < 1.33
Se 5 settlement, (mm)
When the width B is large, the preceding equation can be approximated as
N60
Se
F ¢ ≤
0.08 d 25
Df
N60
Se (mm)
5
B1 1 0.33¢ ≤ R B
R
0.08
B
25
qnet (kN>m2 ) 5
< 16.63N60 B
Se (mm)
R
25
(6.12)
298 Chapter 6: Mat Foundations
In English units, Eq. (6.12) may be expressed as
qnet(all) (kip>ft 2 ) 5 0.25N60 B1 1 0.33¢
< 0.33N60 3Se (in.) 4
Df
B
≤ R 3Se (in.)4
(6.13)
Generally, shallow foundations are designed for a maximum settlement of 25 mm
and a differential settlement of about 19 mm.
However, the width of the raft foundations are larger than those of the isolated spread
footings. As shown in Table 5.3, the depth of significant stress increase in the soil below a
foundation depends on the width of the foundation. Hence, for a raft foundation, the depth
of the zone of influence is likely to be much larger than that of a spread footing. Thus, the
loose soil pockets under a raft may be more evenly distributed, resulting in a smaller differential settlement. Accordingly, the customary assumption is that, for a maximum raft
settlement of 50 mm, the differential settlement would be 19 mm. Using this logic and
conservatively assuming that Fd 5 1, we can respectively approximate Eqs. (6.12) and
(6.13) as
qnet(all) 5 qnet (kN>m2 ) < 25N60
(6.14a)
The net allowable pressure applied on a foundation (see Figure 6.7) may be expressed as
q5
Q
2 gDf
A
where
Q 5 dead weight of the structure and the live load
A 5 area of the raft
In all cases, q should be less than or equal to allowable qnet.
Unit weight Df
Q
Figure 6.7 Definition of net pressure on soil caused by a mat foundation
(6.15)
6.5 Differential Settlement of Mats
299
Example 6.1
Determine the net ultimate bearing capacity of a mat foundation measuring
15 m 3 10 m on a saturated clay with cu 5 95 kN>m2, f 5 0, and Df 5 2 m.
Solution
From Eq. (6.10),
qnet(u) 5 5.14cu B1 1 ¢
Df
0.195B
≤ R B1 1 0.4 R
L
B
5 (5.14) (95) B1 1 ¢
0.195 3 10
0.4 3 2
≤ R B1 1 ¢
≤R
15
10
5 595.9 kN , m2
■
Example 6.2
What will be the net allowable bearing capacity of a mat foundation with dimensions of
15 m 3 10 m constructed over a sand deposit? Here, Df 5 2 m, the allowable settlement
is 25 mm, and the average penetration number N60 5 10.
Solution
From Eq. (6.12),
qnet(all) 5
Df
N60
Se
Se
B1 1 0.33¢ ≤ R ¢ ≤ < 16.63N60 ¢ ≤
0.08
B
25
25
qnet(all) 5
10
0.33 3 2 25
B1 1
R ¢ ≤ 5 133.25 kN , m2
0.08
10
25
or
6.5
■
Differential Settlement of Mats
In 1988, the American Concrete Institute Committee 336 suggested a method for calculating the differential settlement of mat foundations. According to this method, the rigidity
factor Kr is calculated as
ErIb
(6.16)
Kr 5
EsB3
where
Er 5
Es 5
B5
Ib 5
modulus of elasticity of the material used in the structure
modulus of elasticity of the soil
width of foundation
moment of inertia of the structure per unit length at right angles to B
The term ErIb can be expressed as
ah3
ErIb 5 Er ¢IF 1 a Irb 1 a
≤
12
(6.17)
300 Chapter 6: Mat Foundations
where
ErIb 5 flexural rigidity of the superstructure and foundation per unit length
at right angles to B
SErI br 5 flexural rigidity of the framed members at right angles to B
S(Erah3>12) 5 flexural rigidity of the shear walls
a 5 shear wall thickness
h 5 shear wall height
ErIF 5 flexibility of the foundation
Based on the value of Kr , the ratio (d) of the differential settlement to the total settlement
can be estimated in the following manner:
1. If Kr . 0.5, it can be treated as a rigid mat, and d 5 0.
2. If Kr 5 0.5, then d < 0.1.
3. If Kr 5 0, then d 5 0.35 for square mats (B>L 5 1) and d 5 0.5 for long foundations (B>L 5 0).
6.6
Field Settlement Observations for Mat Foundations
Several field settlement observations for mat foundations are currently available in the literature. In this section, we compare the observed settlements for some mat foundations
constructed over granular soil deposits with those obtained from Eqs. (6.12) and (6.13).
Meyerhof (1965) compiled the observed maximum settlements for mat foundations
constructed on sand and gravel, as listed in Table 6.1. In Eq. (6.12), if the depth factor,
1 1 0.33(Df>B), is assumed to be approximately unity, then
Se (mm) <
2qnet(all)
N60
(6.18)
From the values of qnet(all) and N60 given in Columns 6 and 5, respectively, of Table 6.1,
the magnitudes of Se were calculated and are given in Column 8.
Column 9 of Table 6.1 gives the ratios of calculated to measured values of Se . These
ratios vary from about 0.79 to 3.39. Thus, calculating the net allowable bearing capacity
with the use of Eq. (6.12) or (6.13) will yield safe and conservative values.
6.7
Compensated Foundation
Figure 6.7 and Eq. (6.15) indicate that the net pressure increase in the soil under a mat foundation can be reduced by increasing the depth Df of the mat. This approach is generally referred
to as the compensated foundation design and is extremely useful when structures are to be built
on very soft clays. In this design, a deeper basement is made below the higher portion of the
superstructure, so that the net pressure increase in soil at any depth is relatively uniform. (See
Figure 6.8.) From Eq. (6.15) and Figure 6.7, the net average applied pressure on soil is
q5
Q
2 gDf
A
301
T. Edison
São Paulo, Brazil
Banco do Brazil
São Paulo, Brazil
Iparanga
São Paulo, Brazil
C.B.I., Esplanda
São Paulo, Brazil
Riscala
São Paulo, Brazil
Thyssen
Düsseldorf, Germany
Ministry
Düsseldorf, Germany
Chimney
Cologne, Germany
2
3
4
5
6
7
8
Structure
(2)
1
Case
No.
(1)
Schultze (1962)
Schultze (1962)
Schultze (1962)
Vargas (1948)
Vargas (1961)
Vargas (1948)
Rios and Silva (1948);
Vargas (1961)
Rios and Silva (1948)
Reference
(3)
20.42
15.85
22.55
3.96
14.63
9.14
22.86
18.29
B
m
(4)
10
20
25
20
22
9
18
15
Average
N60
(5)
172.4
220.2
239.4
229.8
383.0
304.4
239.4
229.8
qnet(all)
kN , m2
(6)
10.16
20.32
24.13
12.7
27.94
35.56
27.94
15.24
Observed
maximum
settlement,
Se mm
(7)
34.48
22.02
19.15
22.98
34.82
67.64
26.6
30.64
Calculated
maximum
settlement,
Se mm
(8)
3.39
1.08
0.79
1.81
1.25
1.9
0.95
2.01
calculated Se
observed Se
(9)
Table 6.1 Settlement of Mat Foundations on Sand and Gravel (Based on Meyerhof, 1965) (Based on Meyerhof, G. G., (1965). “Shallow Foundations,”
Journal of the Soil Mechanics and Foundation Engineering Divsion, American Society of Civil Engineers, Vol. 91, No. 2, pp. 21–31, Table 1.
With permission from ASCE.)
302 Chapter 6: Mat Foundations
Figure 6.8 Compensated
foundation
For no increase in the net pressure on soil below a mat foundation, q should be zero. Thus,
Df 5
Q
Ag
(6.19)
This relation for Df is usually referred to as the depth of a fully compensated foundation.
The factor of safety against bearing capacity failure for partially compensated foundations (i.e., Df , Q>Ag) may be given as
FS 5
qnet(u)
q
5
qnet(u)
Q
2 gDf
A
(6.20)
where qnet(u) 5 net ultimate bearing capacity.
For saturated clays, the factor of safety against bearing capacity failure can thus be
obtained by substituting Eq. (6.10) into Eq. (6.20):
5.14cu ¢ 1 1
FS 5
Df
0.195B
≤ ¢1 1 0.4 ≤
L
B
Q
2 gDf
A
(6.21)
Example 6.3
The mat shown in Figure 6.7 has dimensions of 18.3 m 3 30.5 m. The total dead
and live load on the mat is 111 3 103 kN kip. The mat is placed over a saturated clay
having a unit weight of 18.87 kN>m3 and cu 5 134 kN>m2. Given that Df 5 1.52 m,
determine the factor of safety against bearing capacity failure.
6.7 Compensated Foundation
303
Solution
From Eq. (6.21), the factor of safety
5.14cu ¢1 1
FS 5
Df
0.195B
≤ ¢ 1 1 0.4 ≤
L
B
Q
2 gDf
A
We are given that cu 5 134 kN>m2, Df 5 1.52 m, B 5 18.3 m, L 5 30.5 m, and g 5
18.87 kN>m3. Hence,
(5.14) (134) B1 1
FS 5
(0.195) (18.3)
1.52
R B1 1 0.4¢
≤R
30.5
18.3
111 3 103 kN
¢
≤ 2 (18.87) (1.52)
18.3 3 30.5
5 4.66
■
Example 6.4
Consider a mat foundation 30 m 3 40 m in plan, as shown in Figure 6.9. The total dead
load and live load on the raft is 200 3 103 kN. Estimate the consolidation settlement at
the center of the foundation.
Solution
From Eq. (1.61)
Sc(p) 5
CcHc
sor 1 Dsav
r
log ¢
≤
1 1 eo
sor
6
sor 5 (3.67)(15.72) 1 (13.33) (19.1 2 9.81) 1 (18.55 2 9.81) < 208 kN>m2
2
Hc 5 6 m
Cc 5 0.28
eo 5 0.9
For Q 5 200 3 103 kN, the net load per unit area is
q5
Q
200 3 103
2 gDf 5
2 (15.72) (2) < 135.2 kN>m2
A
30 3 40
r we refer to Section 5.5. The loaded area can be divided into
In order to calculate Dsav
four areas, each measuring 15 m 3 20 m. Now using Eq. (5.19), we can calculate the
average stress increase in the clay layer below the corner of each rectangular area, or
Dsav(H
r 2>H1) 5 qo B
H2Ia(H2) 2 H1Ia(H1)
5 135.2B
R
H2 2 H1
(1.67 1 13.3 1 6)Ia(H2) 2 (1.67 1 13.33)Ia(H1)
6
R
304 Chapter 6: Mat Foundations
Q
30 m 40 m
2m
1.67 m
Sand
␥ 15.72 kN/m3
Groundwater table
z
Sand
␥sat 19.1 kN/m3
13.33 m
Normally consolidated clay
␥sat 18.55 kN/m3
Cc 0.28; eo 0.9
6m
Sand
Figure 6.9 Consolidation
settlement under a mat
foundation
For Ia(H2) ,
B
15
5
5 0.71
H2
1.67 1 13.33 1 6
L
20
n2 5
5
5 0.95
H2
21
m2 5
From Fig. 5.7, for m2 5 0.71 and n2 5 0.95, the value of Ia(H2) is 0.21. Again, for Ia(H1),
B
15
m2 5
5
51
H1
15
L
20
n2 5
5
5 1.33
H1
15
From Figure 5.7, Ia(H1) 5 0.225, so
Dsav(H
r 2>H1) 5 135.2B
(21) (0.21) 2 (15) (0.225)
R 5 23.32 kN>m2
6
So, the stress increase below the center of the 30 m 3 40 m area is
(4) (23.32) 5 93.28 kN>m2. Thus
Sc(p) 5
6.8
(0.28) (6)
208 1 93.28
log ¢
≤ 5 0.142 m
1 1 0.9
208
5 142 mm
■
Structural Design of Mat Foundations
The structural design of mat foundations can be carried out by two conventional methods:
the conventional rigid method and the approximate flexible method. Finite-difference
and finite-element methods can also be used, but this section covers only the basic concepts of the first two design methods.
6.8 Structural Design of Mat Foundations
305
Conventional Rigid Method
The conventional rigid method of mat foundation design can be explained step by step
with reference to Figure 6.10:
Step 1. Figure 6.10a shows mat dimensions of L 3 B and column loads of Q1 , Q2 ,
Q3 , c. Calculate the total column load as
Q 5 Q1 1 Q2 1 Q3 1 c
(6.22)
Step 2. Determine the pressure on the soil, q, below the mat at points
A, B, C, D, c, by using the equation
q5
Myx
Q
Mxy
6
6
A
Iy
Ix
(6.23)
where
A 5 BL
Ix 5 (1>12)BL3 5 moment of inertia about the x-axis
Iy 5 (1>12)LB3 5 moment of inertia about the y-axis
Mx 5 moment of the column loads about the x-axis 5 Qey
My 5 moment of the column loads about the y-axis 5 Qex
The load eccentricities, ex and ey , in the x and y directions can be determined by using (xr, yr) coordinates:
xr 5
Q1x1r 1 Q2x2r 1 Q3x3r 1 c
Q
(6.24)
and
ex 5 xr 2
B
2
(6.25)
Similarly,
yr 5
Q1y1r 1 Q2y2r 1 Q3y3r 1 c
Q
(6.26)
and
ey 5 yr 2
L
2
(6.27)
Step 3. Compare the values of the soil pressures determined in Step 2 with the net
allowable soil pressure to determine whether q < qall(net) .
Step 4. Divide the mat into several strips in the x and y directions. (See
Figure 6.10). Let the width of any strip be B1 .
y
y
B1
A
B1
B1
B
Q9
B1
D
C
Q11
Q10
Q12
B1
ex
B1
ey
L
E
J
Q5
Q6
Q7
Q8
Q1
Q2
Q3
Q4
H
G
x
B1
I
x
F
B
(a)
FQ1
FQ2
I
FQ4
FQ3
H
G
F
B1 qav(modified)
unit length
B
(b)
Edge
of mat
L
L
d/2
d/2
d/2
L
Edge of
mat
d/2
b o 2L L
Edge of
mat
d/2
d/2
d/2
L
b o L L
(c)
Figure 6.10 Conventional rigid mat foundation design
306
d/2
L
d/2
b o 2(L L)
L
307
6.8 Structural Design of Mat Foundations
Step 5. Draw the shear, V, and the moment, M, diagrams for each individual strip
(in the x and y directions). For example, the average soil pressure of the bottom strip in the x direction of Figure 6.10a is
qav <
qI 1 qF
2
(6.28)
where qI and qF 5 soil pressures at points I and F, as determined from
Step 2.
The total soil reaction is equal to qavB1B. Now obtain the total column
load on the strip as Q1 1 Q2 1 Q3 1 Q4 . The sum of the column loads on
the strip will not equal qavB1B, because the shear between the adjacent
strips has not been taken into account. For this reason, the soil reaction and
the column loads need to be adjusted, or
Average load 5
qavB1B 1 (Q1 1 Q2 1 Q3 1 Q4 )
2
(6.29)
Now, the modified average soil reaction becomes
qav(modified) 5 qav ¢
average load
≤
qavB1B
(6.30)
and the column load modification factor is
F5
average load
Q1 1 Q2 1 Q3 1 Q4
(6.31)
So the modified column loads are FQ1 , FQ2 , FQ3 , and FQ4 . This modified loading on the strip under consideration is shown in Figure 6.10b. The
shear and the moment diagram for this strip can now be drawn, and the procedure is repeated in the x and y directions for all strips.
Step 6. Determine the effective depth d of the mat by checking for diagonal
tension shear near various columns. According to ACI Code 318-95
(Section 11.12.2.1c, American Concrete Institute, 1995), for the critical
section,
U 5 bod3f(0.34)"fcr 4
(6.32)
where
U 5 factored column load (MN), or (column load) 3 (load factor)
f 5 reduction factor 5 0.85
fcr 5 compressive strength of concrete at 28 days (MN>m2 )
The units of bo and d in Eq. (6.32a) are in meters.
The expression for bo in terms of d, which depends on the location
of the column with respect to the plan of the mat, can be obtained from
Figure 6.10c.
308 Chapter 6: Mat Foundations
Step 7. From the moment diagrams of all strips in one direction (x or y), obtain the
maximum positive and negative moments per unit width (i.e., Mr 5 M>B1).
Step 8. Determine the areas of steel per unit width for positive and negative reinforcement in the x and y directions. We have
Mu 5 (Mr) (load factor) 5 fA sfy ¢d 2
a
≤
2
(6.33)
and
a5
A sfy
0.85fcr b
(6.34)
where
As 5
fy 5
Mu 5
f5
area of steel per unit width
yield stress of reinforcement in tension
factored moment
0.9 5 reduction factor
Examples 6.5 and 6.6 illustrate the use of the conventional rigid method of mat foundation
design.
Approximate Flexible Method
In the conventional rigid method of design, the mat is assumed to be infinitely rigid. Also,
the soil pressure is distributed in a straight line, and the centroid of the soil pressure is coincident with the line of action of the resultant column loads. (See Figure 6.11a.) In the
approximate flexible method of design, the soil is assumed to be equivalent to an infinite
number of elastic springs, as shown in Figure 6.11b. This assumption is sometimes
referred to as the Winkler foundation. The elastic constant of these assumed springs is
referred to as the coefficient of subgrade reaction, k.
To understand the fundamental concepts behind flexible foundation design, consider a beam of width B1 having infinite length, as shown in Figure 6.11c. The beam is
subjected to a single concentrated load Q. From the fundamentals of mechanics of
materials,
M 5 EFIF
d2z
dx2
(6.35)
where
M 5 moment at any section
EF 5 modulus of elasticity of foundation material
IF 5 moment of inertia of the cross section of the beam 5 ( 121 )B1h3 (see Figure 6.11c).
However,
dM
5 shear force 5 V
dx
and
dV
5 q 5 soil reaction
dx
6.8 Structural Design of Mat Foundations
309
Q
Q1
Q3
Q2
Resultant of
soil pressure
(a)
Q2
Q1
(b)
Point load
A
B1
x
h
Section
at A A
q
A
z
(c)
Figure 6.11 (a) Principles of design by conventional rigid method; (b) principles of approximate
flexible method; (c) derivation of Eq. (6.39) for beams on elastic foundation
Hence,
d2M
5q
dx2
(6.36)
Combining Eqs. (6.35) and (6.36) yields
EFIF
d4z
5q
dx4
However, the soil reaction is
q 5 2zkr
(6.37)
310 Chapter 6: Mat Foundations
where
z 5 deflection
kr 5 kB1
k 5 coefficient of subgrade reaction (kN>m3 or lb>in3 )
So,
EFIF
d4z
5 2zkB1
dx4
(6.38)
Solving Eq. (6.38) yields
z 5 e2ax (Ar cos bx 1 As sin bx)
(6.39)
where Ar and As are constants and
b5
B1k
Å 4EFIF
4
(6.40)
The unit of the term b, as defined by the preceding equation, is (length) 21. This
parameter is very important in determining whether a mat foundation should be
designed by the conventional rigid method or the approximate flexible method.
According to the American Concrete Institute Committee 336 (1988), mats should be
designed by the conventional rigid method if the spacing of columns in a strip is less
than 1.75>b. If the spacing of columns is larger than 1.75>b, the approximate flexible
method may be used.
To perform the analysis for the structural design of a flexible mat, one must know
the principles involved in evaluating the coefficient of subgrade reaction, k. Before proceeding with the discussion of the approximate flexible design method, let us discuss this
coefficient in more detail.
If a foundation of width B (see Figure 6.12) is subjected to a load per unit area of q, it
will undergo a settlement D. The coefficient of subgrade modulus can be defined as
k5
q
D
(6.41)
B
q
Figure 6.12 Definition of coefficient of subgrade reaction, k
311
6.8 Structural Design of Mat Foundations
The unit of k is kN/m3. The value of the coefficient of subgrade reaction is not a constant
for a given soil, but rather depends on several factors, such as the length L and width B of
the foundation and also the depth of embedment of the foundation. A comprehensive study
by Terzaghi (1955) of the parameters affecting the coefficient of subgrade reaction indicated that the value of the coefficient decreases with the width of the foundation. In the
field, load tests can be carried out by means of square plates measuring 0.3 m 3 0.3 m,
and values of k can be calculated. The value of k can be related to large foundations measuring B 3 B in the following ways:
Foundations on Sandy Soils
For foundations on sandy soils,
k 5 k0.3 ¢
B 1 0.3 2
≤
2B
(6.42)
where k0.3 and k 5 coefficients of subgrade reaction of foundations measuring 0.3 m 3 0.3 m
and B (m) 3 B (m), respectively (unit is kN>m3 ).
Foundations on Clays
For foundations on clays,
k(kN>m3 ) 5 k0.3 (kN>m3 ) B
0.3 (m)
R
B(m)
(6.43)
The definitions of k and k0.3 in Eq. (6.43) are the same as in Eq. (6.42).
For rectangular foundations having dimensions of B 3 L (for similar soil and q),
k(B3B) ¢ 1 1 0.5
k5
1.5
B
≤
L
(6.44)
where
k 5 coefficient of subgrade modulus of the rectangular foundation (L 3 B)
k(B3B) 5 coefficient of subgrade modulus of a square foundation having dimension
of B 3 B
312 Chapter 6: Mat Foundations
Equation (6.44) indicates that the value of k for a very long foundation with a width B is
approximately 0.67k(B3B) .
The modulus of elasticity of granular soils increases with depth. Because the settlement of a foundation depends on the modulus of elasticity, the value of k increases with
the depth of the foundation.
Table 6.2 provides typical ranges of values for the coefficient of subgrade reaction,
k0.3 (k1 ), for sandy and clayey soils.
For long beams, Vesic (1961) proposed an equation for estimating subgrade reaction,
namely,
kr 5 Bk 5 0.65
EsB4 Es
É EFIF 1 2 m2s
12
or
k 5 0.65
EsB4
Es
É EFIF B(1 2 m2s )
12
where
Es 5
B5
EF 5
IF 5
ms 5
modulus of elasticity of soil
foundation width
modulus of elasticity of foundation material
moment of inertia of the cross section of the foundation
Poisson’s ratio of soil
Table 6.2 Typical Subgrade Reaction
Values, k0.3 (k1 )
Soil type
Dry or moist sand:
Loose
Medium
Dense
Saturated sand:
Loose
Medium
Dense
Clay:
Stiff
Very stiff
Hard
k0.3(k1)
MN , m3
8–25
25–125
125–375
10–15
35–40
130–150
10–25
25–50
.50
(6.45)
6.8 Structural Design of Mat Foundations
313
For most practical purposes, Eq. (6.46) can be approximated as
k5
Es
B(1 2 m2s )
(6.46)
Now that we have discussed the coefficient of subgrade reaction, we will proceed
with the discussion of the approximate flexible method of designing mat foundations. This
method, as proposed by the American Concrete Institute Committee 336 (1988), is
described step by step. The use of the design procedure, which is based primarily on the
theory of plates, allows the effects (i.e., moment, shear, and deflection) of a concentrated
column load in the area surrounding it to be evaluated. If the zones of influence of two or
more columns overlap, superposition can be employed to obtain the net moment, shear,
and deflection at any point. The method is as follows:
Step 1. Assume a thickness h for the mat, according to Step 6 of the conventional
rigid method. (Note: h is the total thickness of the mat.)
Step 2. Determine the flexural ridigity R of the mat as given by the formula
R5
EFh3
12(1 2 m2F )
(6.47)
where
EF 5 modulus of elasticity of foundation material
mF 5 Poisson’s ratio of foundation material
Step 3. Determine the radius of effective stiffness—that is,
Lr 5
R
Åk
4
(6.48)
where k 5 coefficient of subgrade reaction. The zone of influence of any
column load will be on the order of 3 to 4 Lr.
Step 4. Determine the moment (in polar coordinates at a point) caused by a column
load (see Figure 6.13a). The formulas to use are
Mr 5 radial moment 5 2
Q
(1 2 mF )A 2
A 2
S
r
4C 1
Lr
(6.49)
and
Mt 5 tangential moment 5 2
Q
(1 2 mF )A 2
mFA 1 1
C
S
r
4
Lr
(6.50)
314 Chapter 6: Mat Foundations
6
5
y
My
Mr
4
Mt
Mx
r
r
3
L
␣
A2
x
A4
2
A1
(a)
A3
1
0
–0.4
–0.3
–0.2
–0.1
0
0.1
A1, A2, A3, A4
(b)
0.2
0.3
0.4
Figure 6.13 Approximate flexible method of mat design
where
r 5 radial distance from the column load
Q 5 column load
A 1 , A 2 5 functions of r>Lr
The variations of A 1 and A 2 with r>Lr are shown in Figure 6.13b. (For
details see Hetenyi, 1946.)
In the Cartesian coordinate system (see Figure 6.13a),
Mx 5 Mt sin2 a 1 Mr cos2 a
(6.51)
My 5 Mt cos2 a 1 Mr sin2 a
(6.52)
and
Step 5. For the unit width of the mat, determine the shear force V caused by a column
load:
Q
V5
A3
(6.53)
4Lr
The variation of A 3 with r>Lr is shown in Figure 6.13b.
Step 6. If the edge of the mat is located in the zone of influence of a column, determine the moment and shear along the edge. (Assume that the mat is continuous.) Moment and shear opposite in sign to those determined are applied
at the edges to satisfy the known conditions.
Step 7. The deflection at any point is given by
d5
QLr2
A
4R 4
The variation of A 4 is presented in Figure 6.13b.
(6.54)
6.8 Structural Design of Mat Foundations
315
Example 6.5
The plan of a mat foundation is shown in Figure 6.14. Calculate the soil pressure at points
A, B, C, D, E, and F. (Note: All column sections are planned to be 0.5 m ⫻ 0.5 m.)
y
y
A
G
B
I
C
0.25 m
400 kN
500 kN
450 kN
7m
1500 kN
1500 kN
1200 kN
4.25 m
8m
4.25 m
7m
x
1500 kN
1500 kN
1200 kN
7m
400 kN
500 kN
350 kN
0.25 m
F
H
E
8m
J
D
x
8m
0.25 m
0.25 m
Figure 6.14 Plan of a mat foundation
Solution
Eq. (6.23): q 5
My x
Q
Mx y
6
6
A
Iy
Ix
A ⫽ (16.5)(21.5) ⫽ 354.75 m2
1
1
Ix 5 BL3 5 (16.5) (21.5) 3 5 13,665 m4
12
12
1
1
3
Iy 5 LB 5 (21.5) (16.5) 3 5 8,050 m4
12
12
Q ⫽ 350 ⫹ (2)(400) ⫹ 450 ⫹ (2)(500) ⫹ (2)(1200) ⫹ (4)(1500) ⫽ 11,000 kN
B
My 5 Qex; ex 5 x9 2
2
316 Chapter 6: Mat Foundations
Q1x1r 1 Q2x2r 1 Q3x3r 1 c
Q
(8.25) (500 1 1500 1 1500 1 500)
1
5
C 1 (16.25) (350 1 1200 1 1200 1 450) S 5 7.814 m
11,000
1 (0.25) (400 1 1500 1 1500 1 400)
B
ex 5 x9 2
5 7.814 2 8.25 5 20.435 m < 20.44 m
2
Hence, the resultant line of action is located to the left of the center of the mat. So
My ⫽ (11,000)(0.44) ⫽ 4840 kN-m. Similarly
L
Mx 5 Qey; ey 5 yr 2
2
Q1y1r 1 Q2y2r 1 Q3y3r 1 c
yr 5
Q
1
(0.25) (400 1 500 1 350) 1 (7.25) (1500 1 1500 1 1200)
5
c
d
11,000 1(14.25) (1500 1 1500 1 1200) 1 (21.25) (400 1 500 1 450)
5 10.85 m
L
ey 5 y9 2 5 10.85 2 10.75 5 0.1m
2
The location of the line of action of the resultant column loads is shown in Figure 6.15.
Mx ⫽ (11,000)(0.1) ⫽ 1100 kN-m. So
11,000
1100y
4840x
q5
6
6
5 31.0 6 0.6x 6 0.08y( kN>m2 )
354.75
8050
13,665
xr 5
y
y
A
G
B
I
C
0.25 m
400 kN
500 kN
450 kN
7m
1500 kN
4.25 m
0.1 m
1500 kN
1500 kN 1200 kN
8m
4.25
m
0.44 m
7m
x
1200 kN
1500 kN
7m
400 kN
500 kN
350 kN
0.25 m
F
H
8m
0.25 m
Figure 6.15
E
J
8m
D
0.25 m
x
6.8 Structural Design of Mat Foundations
Therefore,
At A: q ⫽ 31.0 ⫹ (0.6)(8.25) ⫹ (0.08)(10.75) ⫽ 36.81 kN/m2
At B: q ⫽ 31.0 ⫹ (0.6)(0) ⫹ (0.08)(10.75) ⫽ 31.86 kN/m2
At C: q ⫽ 31.0 ⫺ (0.6)(8.25) ⫹ (0.08)(10.75) ⫽ 26.91 kN/m2
At D: q ⫽ 31.0 ⫺ (0.6)(8.25) ⫺ (0.08)(10.75) ⫽ 25.19 kN/m2
At E: q ⫽ 31.0 ⫹ (0.6)(0) ⫺ (0.08)(10.75) ⫽ 30.14 kN/m2
At F: q ⫽ 31.0 ⫹ (0.6)(8.25) ⫺ (0.08)(10.75) ⫽ 35.09 kN/m2
317
■
Example 6.6
Divide the mat shown in Figure 6.14 into three strips, such as AGHF (B1 ⫽ 4.25 m),
GIJH (B1 ⫽ 8 m), and ICDJ (B1 ⫽ 4.25 m). Use the result of Example 6.5, and determine the reinforcement requirements in the y direction. Here, fc⬘ ⫽ 20.7 MN/m2, fy ⫽
413.7 MN/m2, and the load factor is 1.7.
Solution
Determination of Shear and Moment Diagrams for Strips:
Strip AGHF:
Average soil pressure 5 qav 5 q(at A) 1 q(at F) 5
36.81 1 35.09
5 35.95 kN>m2
2
Total soil reaction ⫽ qav B1L ⫽ (35.95)(4.25)(21.50) ⫽ 3285 kN
load due to soil reaction 1 column loads
Average load 5
2
3285 1 3800
5
5 3542.5 kN
2
So, modified average soil pressure,
qav(modified) 5 qav a
3542.5
3542.5
b 5 (35.95) a
b 5 38.768 kN>m2
3285
3285
The column loads can be modified in a similar manner by multiplying factor
F5
3542.5
5 0.9322
3800
Figure 6.16 shows the loading on the strip and corresponding shear and moment diagrams.
Note that the column loads shown in this figure have been multiplied by F ⫽ 0.9322. Also the
load per unit length of the beam is equal to B1qav(modified) ⫽ (4.25)(38.768) ⫽ 164.76 kN/m.
Strip GIJH: In a similar manner,
q(at B) 1 q(at E)
31.86 1 30.14
5
5 31.0 kN>m2
2
2
Total soil reaction ⫽ (31)(8)(21.5) ⫽ 5332 kN
Total column load ⫽ 4000 kN
qav 5
318 Chapter 6: Mat Foundations
372.89 kN
0.25 m
1398.36 kN
1398.36 kN
7m
7m
372.89 kN
0.25 m
7m
F
A
164.76 kN/m
821.65
576.7
331.7
41.19
Shear (unit: kN)
41.19
381.7
576.70
821.65
1727.57
1127.57
5.15
5.15
718.35
Moment (unit: kN-m)
326.55
326.55
Figure 6.16 Load, shear, and moment diagrams for strip AGHF
Average load 5
5332 1 4000
5 4666 kN
2
q av(modified) 5 (31.0) a
F5
4666
b 5 27.12 kN>m2
5332
4666
5 1.1665
4000
The load, shear, and moment diagrams are shown in Figure 6.17.
Strip ICDJ: Figure 6.18 shows the load, shear, and moment diagrams for this strip.
Determination of the Thickness of the Mat
For this problem, the critical section for diagonal tension shear will be at the column
carrying 1500 kN of load at the edge of the mat [Figure 6.19]. So
bo 5 a0.5 1
d
d
b 1 a0.5 1 b 1 (0.5 1 d) 5 1.5 1 2d
2
2
U 5 (bod) S (f) (0.34)"fcr T
or
U ⫽ (1.7)(1500) ⫽ 2550 kN ⫽ 2.55 MN
2.55 5 (1.5 1 2d) (d) S(0.85) (0.34) !20.7 T
(1.5 ⫹ 2d)(d) ⫽ 1.94; d ⫽ 0.68 m
6.8 Structural Design of Mat Foundations
583.25 kN
0.25 m
1749.75 kN
1749.75 kN
7m
7m
583.25 kN
0.25 m
7m
E
B
217 kN/m
990.17
759.58
529
54.256
Shear (unit: kN)
54.256
529
759.58
990.17
1620.89
1620.89
291.62
6.75
6.78
Moment (unit: kN-m)
637.94
637.94
Figure 6.17 Load, shear, and moment diagrams for strip GIJH
1046.44 kN
1046.44 kN
392.4 kN
0.25 m
7m
7m
305.2 kN
0.25 m
7m
D
C
129.8 kN/m
548.65
410.81
272.97
32.45
Shear (unit: kN)
32.3
359.95
497.79
635.63
664.56
360.15
4.06
4.06
Moment (unit: kN-m)
289.95
495.0
1196.19
Figure 6.18 Load, shear, and moment diagrams for strip ICDJ
319
320 Chapter 6: Mat Foundations
1500 kN
Column load
Edge
of mat
0.5 + d
0.5 + d/2
Figure 6.19 Critical perimeter column
Assuming a minimum cover of 76 mm over the steel reinforcement and also assuming that the steel bars to be used are 25 mm in diameter, the total thickness of the
slab is
h ⫽ 0.68 ⫹ 0.076 ⫹ 0.025 ⫽ 0.781 m ⬇ 0.8 m
The thickness of this mat will satisfy the wide beam shear condition across the three
strips under consideration.
Determination of Reinforcement
From the moment diagram shown in Figures 6.16, 6.17, and 6.18, it can be seen that
the maximum positive moment is located in strip AGHF, and its magnitude is
M9 5
1727.57
1727.57
5
5 406.5 kN-m>m
B1
4.25
Similarly, the maximum negative moment is located in strip ICDJ and its magnitude is
Mr 5
1196.19
1196.19
5
5 281.5 kN-m>m
B1
4.25
a
From Eq. (6.33): Mu 5 (M r ) ( load factor) 5 fA s fy ad 2 b .
2
a
For the positive moment, Mu 5 (406.5)(1.7) 5 (f)(As ) (413.7 3 1000) a0.68 2 b
2
␾ ⫽ 0.9. Also, from Eq. (6.34),
As fy
(As )(413.7)
5 23.51As; or As 5 0.0425a
(0.85) (20.7) (1)
a
691.05 5 (0.9)(0.0425a) (413,700) a0.68 2 b; or a < 0.0645
2
a5
0.85 f9c b
5
So, As ⫽ (0.0425)(0.0645) ⫽ 0.00274 m2/m ⫽ 2740 mm2/m.
6.8 Structural Design of Mat Foundations
321
Use 25-mm diameter bars at 175 mm center-to-center:
cA s provided 5 (491) a
1000
b 5 2805.7 mm2>md
175
Similarly, for negative reinforcement,
a
Mu 5 (281.5) (1.7) 5 (f)(As )(413.7 3 1000) a0.68 2 b
2
␾ ⫽ 0.9. As ⫽ 0.0425a
So
a
478.55 5 (0.9) (0.0425a) (413.7 3 1000) a0.68 2 b; or a < 0.045
2
So, As ⫽ (0.045)(0.0425) ⫽ 0.001913 m2/m ⫽ 1913 mm2/m.
Use 25-mm diameter bars at 255 mm center-to-center:
[As provided ⫽ 1925 mm2]
Because negative moment occurs at midbay of strip ICDJ, reinforcement should be provided. This moment is
M9 5
289.95
5 68.22 kN- m>m
4.25
Hence,
a
Mu 5 (68.22)(1.7) 5 (0.9)(0.0425a)(413.7 3 1000) a0.68 2 b;
2
or a < 0.0108
As ⫽ (0.0108)(0.0425) ⫽ 0.000459 m2/m ⫽ 459 mm2/m
Provide 16-mm diameter bars at 400 mm center-to-center:
[As provided ⫽ 502 mm2]
For general arrangement of the reinforcement see Figure 6.20.
Top steel
Bottom steel
Additional top steel
in strip ICDJ
Top steel
Figure 6.20 General arrangement of reinforcement
■
322 Chapter 6: Mat Foundations
Problems
6.1
Determine the net ultimate bearing capacity of mat foundations with the following
characteristics:
cu 5 120 kN>m2, f 5 0, B 5 8 m, L 5 18 m, Df 5 3 m
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
Following are the results of a standard penetration test in the field (sandy soil):
Depth (m)
Field value of N60
1.5
3.0
4.5
6.0
7.5
9.0
10.5
9
12
11
7
13
11
13
Estimate the net allowable bearing capacity of a mat foundation 6.5 m 3 5 m in
plan. Here, Df 5 1.5 m and allowable settlement 5 50 mm. Assume that the unit
weight of soil, g 5 16.5 kN>m3.
Repeat Problem 6.2 for an allowable settlement of 30 mm.
A mat foundation on a saturated clay soil has dimensions of 20 m 3 20 m. Given:
dead and live load 5 48 MN, cu 5 30 kN>m2, and gclay 5 18.5 kN>m3.
a. Find the depth, Df, of the mat for a fully compensated foundation.
b. What will be the depth of the mat (Df ) for a factor of safety of 2 against bearing
capacity failure?
Repeat Problem 6.4 part b for cu 5 20 kN>m2 .
A mat foundation is shown in Figure P6.6. The design considerations
are L 5 12 m, B 5 10 m, Df 5 2.2 m, Q 5 30 MN, x1 5 2 m, x2 5 2 m,
x3 5 5.2 m, and preconsolidation pressure scr < 105 kN>m 2. Calculate the consolidation settlement under the center of the mat.
For the mat foundation in Problem 6.6, estimate the consolidation settlement under
the corner of the mat.
From the plate load test (plate dimensions 0.3 m 3 0.3 m) in the field, the coefficient of subgrade reaction of a sandy soil is determined to be 14,900 kN>m3. What
will be the value of the coefficient of subgrade reaction on the same soil for a foundation with dimensions of 7.5 m 3 7.5 m?
Refer to Problem 6.18. If the full-sized foundation had dimensions of 21.3 m 3 9.1 m,
what will be the value of the coefficient of subgrade reaction?
The subgrade reaction of a sandy soil obtained from the plate load test (plate
dimensions 1 m 3 0.7 m ) is 18 MN>m3. What will be the value of k on the same
soil for a foundation measuring 5 m 3 3.5 m ?
References
323
Size of mat B L
Sand
␥ 16.0 kN/m3
Df
Q
x1
z
x2
x3
Groundwater
table
Sand
␥ sat 18.0 kN/m3
Clay
␥sat 17.5 kN/m3
eo 0.88
Cc 0.38
Cs 0.1
Figure P6.6
References
AMERICAN CONCRETE INSTITUTE (1995). ACI Standard Building Code Requirements for Reinforced Concrete, ACI 318–95, Farmington Hills, MI.
AMERICAN CONCRETE INSTITUTE COMMITTEE 336 (1988). “Suggested Design Procedures for
Combined Footings and Mats,” Journal of the American Concrete Institute, Vol. 63, No. 10,
pp. 1041–1077.
HETENYI, M. (1946). Beams of Elastic Foundations, University of Michigan Press, Ann Arbor, MI.
MEYERHOF, G. G. (1965). “Shallow Foundations,” Journal of the Soil Mechanics and Foundations
Division, American Society of Civil Engineers, Vol. 91, No. SM2, pp. 21–31.
RIOS, L., and SILVA, F. P. (1948). “Foundations in Downtown São Paulo (Brazil),” Proceedings,
Second International Conference on Soil Mechanics and Foundation Engineering, Rotterdam,
Vol. 4, p. 69.
SCHULTZE, E. (1962). “Probleme bei der Auswertung von Setzungsmessungen,” Proceedings,
Baugrundtagung, Essen, Germany, p. 343.
TERZAGHI, K. (1955). “Evaluation of the Coefficient of Subgrade Reactions,” Geotechnique,
Institute of Engineers, London, Vol. 5, No. 4, pp. 197–226.
VARGAS, M. (1948). “Building Settlement Observations in São Paulo,” Proceedings Second International Conference on Soil Mechanics and Foundation Engineering, Rotterdam, Vol. 4,
p. 13.
VARGAS, M. (1961). “Foundations of Tall Buildings on Sand in São Paulo (Brazil),” Proceedings,
Fifth International Conference on Soil Mechanics and Foundation Engineering, Paris, Vol. 1,
p. 841.
VESIC, A. S. (1961). “Bending of Beams Resting on Isotropic Solid,” Journal of the Engineering
Mechanics Division, American Society of Civil Engineers, Vol. 87, No. EM2, pp. 35–53.
7
7.1
Lateral Earth Pressure
Introduction
Vertical or near-vertical slopes of soil are supported by retaining walls, cantilever sheet-pile
walls, sheet-pile bulkheads, braced cuts, and other, similar structures. The proper design of
those structures requires an estimation of lateral earth pressure, which is a function of several factors, such as (a) the type and amount of wall movement, (b) the shear strength parameters of the soil, (c) the unit weight of the soil, and (d) the drainage conditions in the
backfill. Figure 7.1 shows a retaining wall of height H. For similar types of backfill,
a. The wall may be restrained from moving (Figure 7.1a). The lateral earth pressure on
the wall at any depth is called the at-rest earth pressure.
b. The wall may tilt away from the soil that is retained (Figure 7.1b). With sufficient
wall tilt, a triangular soil wedge behind the wall will fail. The lateral pressure for
this condition is referred to as active earth pressure.
c. The wall may be pushed into the soil that is retained (Figure 7.1c). With sufficient wall
movement, a soil wedge will fail. The lateral pressure for this condition is referred to
as passive earth pressure.
Figure 7.2 shows the nature of variation of the lateral pressure, shr , at a certain depth of the
wall with the magnitude of wall movement.
+ H
– H
h (at-rest)
h (active)
h (passive)
Soil
failure
wedge
Height = H
Height = H
(a)
324
Soil
failure
wedge
Height = H
(b)
Figure 7.1 Nature of lateral earth pressure on a retaining wall
(c)
7.2 Lateral Earth Pressure at Rest
325
h
h (passive)
( )
H
H
p
( )
H
H
⬇ 0.01 for loose
sand to 0.05
for soft clay
( )
H
H
H
H
h (active)
p
a
⬇ 0.001 for loose
sand to 0.04
for soft clay
h (at rest)
( )
H
H
a
H
H
Figure 7.2 Nature of variation of lateral earth pressure at a certain depth
In the sections that follow, we will discuss various relationships to determine
the at-rest, active, and passive pressures on a retaining wall. It is assumed that the reader
has studied lateral earth pressure in the past, so this chapter will serve as a review.
7.2
Lateral Earth Pressure at Rest
Consider a vertical wall of height H, as shown in Figure 7.3, retaining a soil having a unit
weight of g. A uniformly distributed load, q>unit area, is also applied at the ground surface. The shear strength of the soil is
s 5 cr 1 sr tan fr
where
cr 5 cohesion
fr 5 effective angle of friction
sr 5 effective normal stress
q
Ko q
o
c
z
H
1
2
P1
Po
h
P2
H/2
H/3
Ko (q + H)
(a)
Figure 7.3 At-rest earth pressure
(b)
z
326 Chapter 7: Lateral Earth Pressure
At any depth z below the ground surface, the vertical subsurface stress is
(7.1)
sor 5 q 1 gz
If the wall is at rest and is not allowed to move at all, either away from the soil mass
or into the soil mass (i.e., there is zero horizontal strain), the lateral pressure at a depth z is
(7.2)
sh 5 Kosor 1 u
where
u 5 pore water pressure
Ko 5 coefficient of at-rest earth pressure
For normally consolidated soil, the relation for Ko (Jaky, 1944) is
Ko < 1 2 sin fr
(7.3)
Equation (7.3) is an empirical approximation.
For overconsolidated soil, the at-rest earth pressure coefficient may be expressed as
(Mayne and Kulhawy, 1982)
Ko 5 (1 2 sin fr)OCRsin fr
(7.4)
where OCR 5 overconsolidation ratio.
With a properly selected value of the at-rest earth pressure coefficient, Eq. (7.2) can
be used to determine the variation of lateral earth pressure with depth z. Figure 7.3b shows
the variation of shr with depth for the wall depicted in Figure 7.3a. Note that if the surcharge q 5 0 and the pore water pressure u 5 0, the pressure diagram will be a triangle.
The total force, Po , per unit length of the wall given in Figure 7.3a can now be obtained
from the area of the pressure diagram given in Figure 7.3b and is
Po 5 P1 1 P2 5 qKoH 1 12gH 2Ko
(7.5)
where
P1 5 area of rectangle 1
P2 5 area of triangle 2
The location of the line of action of the resultant force, Po , can be obtained by taking the
moment about the bottom of the wall. Thus,
P1 ¢
z5
H
H
≤ 1 P2 ¢ ≤
2
3
Po
(7.6)
If the water table is located at a depth z , H, the at-rest pressure diagram shown in
Figure 7.3b will have to be somewhat modified, as shown in Figure 7.4. If the effective unit
weight of soil below the water table equals gr (i.e., gsat 2 gw ), then
at z 5 0,
at z 5 H1 ,
shr 5 Kosor 5 Koq
shr 5 Kosor 5 Ko (q 1 gH1 )
and
at z 5 H2 ,
shr 5 Kosor 5 Ko (q 1 gH1 1 grH2 )
7.2 Lateral Earth Pressure at Rest
327
q
Ko q
c
H1
1
z
2
Water table
Ko (q + H1 )
H
3
sat
c
H2
h
u
5
4
Ko (q + H1 + H2 )
(a)
w H2
(b)
Figure 7.4 At-rest earth pressure with water table located at a depth z , H
Note that in the preceding equations, sor and shr are effective vertical and horizontal pressures, respectively. Determining the total pressure distribution on the wall requires adding
the hydrostatic pressure u, which is zero from z 5 0 to z 5 H1 and is H2gw at z 5 H2 . The
variation of shr and u with depth is shown in Figure 7.4b. Hence, the total force per unit
length of the wall can be determined from the area of the pressure diagram. Specifically,
Po 5 A 1 1 A 2 1 A 3 1 A 4 1 A 5
where A 5 area of the pressure diagram.
So,
Po 5 KoqH1 1 12KogH 21 1 Ko (q 1 gH1 )H2 1 12KogrH 22 1 12gwH 22
(7.7)
Example 7.1
For the retaining wall shown in Figure 7.5(a), determine the lateral earth force at rest per unit
length of the wall. Also determine the location of the resultant force. Assume OCR ⫽ 1.
Solution
Ko ⫽ 1 ⫺ sin ␾⬘ ⫽ 1 ⫺ sin 30° ⫽ 0.5
At z ⫽ 0, ␴o⬘ ⫽ 0; ␴h⬘ ⫽ 0
At z ⫽ 2.5 m, ␴o⬘ ⫽ (16.5)(2.5) ⫽ 41.25 kN/m2;
␴h⬘ ⫽ Ko␴o⬘ ⫽ (0.5)(41.25) ⫽ 20.63 kN/m2
At z ⫽ 5 m, ␴o⬘ ⫽ (16.5)(2.5) ⫹ (19.3 ⫺ 9.81)2.5 ⫽ 64.98 kN/m2;
␴h⬘ ⫽ Ko␴o⬘ ⫽ (0.5)(64.98) ⫽ 32.49 kN/m2
328 Chapter 7: Lateral Earth Pressure
= 16.5 kN/m3
= 30
c = 0
Ground
water table
2.5 m
z
sat = 19.3 kN/m3
= 30
c = 0
2.5 m
1
20.63 kN/m2
2
h
3
O
(a)
32.49 kN/m2
u
4
24.53
kN/m2
(b)
Figure 7.5
The hydrostatic pressure distribution is as follows:
From z ⫽ 0 to z ⫽ 2.5 m, u ⫽ 0. At z ⫽ 5 m, u ⫽ ␥w(2.5) ⫽ (9.81)(2.5) ⫽ 24.53
kN/m2. The pressure distribution for the wall is shown in Figure 7.5b.
The total force per unit length of the wall can be determined from the area of the
pressure diagram, or
Po 5 Area 1 1 Area 2 1 Area 3 1 Area 4
5 12 (2.5) (20.63) 1 (2.5) (20.63) 1 12 (2.5) (32.49 2 20.63)
1 12 (2.5) (24.53) 5 122.85 kN>m
The location of the center of pressure measured from the bottom of the wall (point O) ⫽
(
z5
Area 1)a2.51
2.5
3
b1( Area 2)a
2.5
2
b1( Area 31 Area 4)a
2.5
3
b
Po
(25.788) (3.33) 1 (51.575) (1.25) 1 (14.825 1 30.663) (0.833)
122.85
85.87 1 64.47 1 37.89
5
5 1.53 m
122.85
5
■
Active Pressure
7.3
Rankine Active Earth Pressure
The lateral earth pressure described in Section 7.2 involves walls that do not yield at all.
However, if a wall tends to move away from the soil a distance Dx, as shown in Figure 7.6a,
the soil pressure on the wall at any depth will decrease. For a wall that is frictionless, the
horizontal stress, shr , at depth z will equal Kosro (5Kogz) when Dx is zero. However, with
Dx . 0, shr will be less than Kosor .
Wall
movement
to left
45 /2
x
45 /2
z
z
o
c
H
h
Rotation of wall
about this point
(a)
Shear
stress
s
c
ta
c
b
n
a
a
h
Koo
o
Effective
normal
stress
(b)
2c Ka
zc
H
o Ka
2c Ka
Pa
Eq. (7.12)
o Ka 2c Ka
(c)
Figure 7.6 Rankine active pressure
329
330 Chapter 7: Lateral Earth Pressure
The Mohr’s circles corresponding to wall displacements of Dx 5 0 and Dx . 0 are
shown as circles a and b, respectively, in Figure 7.6b. If the displacement of the wall, Dx,
continues to increase, the corresponding Mohr’s circle eventually will just touch the
Mohr–Coulomb failure envelope defined by the equation
s 5 cr 1 sr tan fr
This circle, marked c in the figure, represents the failure condition in the soil mass; the horizontal stress then equals sar , referred to as the Rankine active pressure. The slip lines (failure planes) in the soil mass will then make angles of 6(45 1 fr>2) with the horizontal, as
shown in Figure 7.6a.
Equation (1.87) relates the principal stresses for a Mohr’s circle that touches the
Mohr–Coulomb failure envelope:
s1r 5 s3r tan2 ¢ 45 1
fr
fr
≤ 1 2cr tan ¢ 45 1 ≤
2
2
For the Mohr’s circle c in Figure 7.6b,
Major principal stress, s1r 5 sor
and
Minor principal stress, s3r 5 sar
Thus,
fr
fr
≤ 1 2cr tan ¢45 1 ≤
2
2
sor
2cr
2
sar 5
fr
fr
2
tan ¢45 1 ≤
tan ¢45 1 ≤
2
2
sor 5 sar tan2 ¢ 45 1
or
sar 5 sor tan2 ¢ 45 2
fr
fr
≤ 2 2cr tan ¢45 2 ≤
2
2
5 sor Ka 2 2cr"Ka
(7.8)
where Ka 5 tan2 (45 2 fr>2) 5 Rankine active-pressure coefficient.
The variation of the active pressure with depth for the wall shown in Figure 7.6a is given
in Figure 7.6c. Note that sor 5 0 at z 5 0 and sor 5 gH at z 5 H. The pressure distribution
shows that at z 5 0 the active pressure equals 22cr"Ka , indicating a tensile stress that
decreases with depth and becomes zero at a depth z 5 zc , or
gzcKa 2 2cr"Ka 5 0
7.3 Rankine Active Earth Pressure
331
and
zc 5
2cr
(7.9)
g"Ka
The depth zc is usually referred to as the depth of tensile crack, because the tensile
stress in the soil will eventually cause a crack along the soil–wall interface. Thus, the total
Rankine active force per unit length of the wall before the tensile crack occurs is
H
H
H
Pa 5 3 sar dz 5 3 gzKa dz 2 3 2cr"Ka dz
0
0
0
5 12gH 2Ka 2 2crH"Ka
(7.10)
After the tensile crack appears, the force per unit length on the wall will be
caused only by the pressure distribution between depths z 5 zc and z 5 H, as shown
by the hatched area in Figure 7.6c. This force may be expressed as
Pa 5 12 (H 2 zc ) (gHKa 2 2cr"Ka )
(7.11)
1
2cr
≤ agHKa 2 2cr"Ka b
Pa 5 ¢ H 2
2
g"Ka
(7.12)
or
However, it is important to realize that the active earth pressure condition will be reached
only if the wall is allowed to “yield” sufficiently. The necessary amount of outward displacement of the wall is about 0.001H to 0.004H for granular soil backfills and about
0.01H to 0.04H for cohesive soil backfills.
Note further that if the total stress shear strength parameters (c, f) were used, an
equation similar to Eq. (7.8) could have been derived, namely,
sa 5 so tan2 ¢ 45 2
f
f
≤ 2 2c tan ¢45 2 ≤
2
2
Example 7.2
A 6-m-high retaining wall is to support a soil with unit weight ␥ ⫽ 17.4 kN/m3, soil
friction angle ␾⬘ ⫽ 26°, and cohesion c⬘ ⫽ 14.36 kN/m2. Determine the Rankine active
force per unit length of the wall both before and after the tensile crack occurs, and
determine the line of action of the resultant in both cases.
Solution
For ␾⬘ ⫽ 26°,
Ka 5 tan2 a45 2
!Ka 5 0.625
fr
b 5 tan2 (45 2 13) 5 0.39
2
sar 5 gHKa 2 2cr !Ka
332 Chapter 7: Lateral Earth Pressure
From Figure 7.6c, at z ⫽ 0,
sar 5 22cr !Ka 5 22(14.36) (0.625) 5 217.95 kN>m2
and at z ⫽ 6 m,
␴a⬘ ⫽ (17.4)(6)(0.39) ⫺ 2(14.36)(0.625)
⫽ 40.72 ⫺ 17.95 ⫽ 22.77 kN/m2
Active Force before the Tensile Crack Appeared: Eq. (7.10)
Pa 5 12 gH 2Ka 2 2crH!Ka
5 12 (6) (40.72) 2 (6) (17.95) 5 122.16 2 107.7 5 14.46 kN>m
The line of action of the resultant can be determined by taking the moment of the area
of the pressure diagrams about the bottom of the wall, or
6
6
Paz 5 (122.16) a b 2 (107.7) a b
3
2
Thus,
z5
244.32 2 323.1
5 25.45 m.
14.46
Active Force after the Tensile Crack Appeared: Eq. (7.9)
zc 5
2(14.36)
2cr
5
5 2.64 m
(17.4) (0.625)
g !Ka
Using Eq. (7.11) gives
Pa 5 12 (H 2 zc ) (gHKa 2 2c9 !Ka ) 5 12 (6 2 2.64) (22.77) 5 38.25 kN>m
Figure 7.6c indicates that the force Pa ⫽ 38.25 kN/m is the area of the hatched triangle.
Hence, the line of action of the resultant will be located at a height –z ⫽ (H ⫺ zc)/3
above the bottom of the wall, or
z5
6 2 2.64
5 1.12 m
3
■
Example 7.3
Assume that the retaining wall shown in Figure 7.7a can yield sufficiently to develop an
active state. Determine the Rankine active force per unit length of the wall and the location of the resultant line of action.
Solution
If the cohesion, c⬘, is zero, then
␴a⬘ ⫽ ␴o⬘Ka
7.3 Rankine Active Earth Pressure
333
For the top layer of soil, ␾1⬘ ⫽ 30°, so
Ka(1) 5 tan2 a45 2
fr1
1
b 5 tan2 (45 2 15) 5
2
3
Similarly, for the bottom layer of soil, ␾2⬘ ⫽ 36°, and it follows that
Ka(2) 5 tan2 a45 2
36
b 5 0.26
2
The following table shows the calculation of ␴a⬘ and u at various depths below the
ground surface.
Depth, z
(m)
␴o⬘
(lb/ft2)
Ka
␴a⬘ ⫽ Ka␴o⬘
(lb/ft2)
u
(lb/ft2)
0
3.05⫺
3.05⫹
6.1
0
(16)(3.05) ⫽ 48.8
48.8
(16)(3.05) ⫹ (19 ⫺ 9.81)(3.05) ⫽ 76.83
1/3
1/3
0.26
0.26
0
16.27
12.69
19.98
0
0
0
(9.81)(3.05) ⫽ 29.92
The pressure distribution diagram is plotted in Figure 7.7b. The force per unit length is
Pa ⫽ area 1 ⫹ area 2 ⫹ area 3 ⫹ area 4
5 12 (3.05)(16.27) 1 (12.69)(3.05) 1 12 (19.98 2 12.69)(3.05) 1 12 (29.92)(3.05)
⫽ 24.81 ⫹ 38.70 ⫹ 11.12 ⫹ 45.63 ⫽ 120.26 kN/m
= 16 kN/m3
′1 = 30°
c′1 = 0
3.05 m
z
Water table
sat = 19 kN/m3
′2 = 36°
c′2 = 0
3.05 m
1
12.69
kN/m2
16.27 kN/m2
2
3
4
19.98
kN/m2
0
(a)
Figure 7.7 Rankine active force behind a retaining wall
29.92
kN/m2
(b)
■
334
Chapter 7: Lateral Earth Pressure
The distance of the line of action of the resultant force from the bottom of the wall
can be determined by taking the moments about the bottom of the wall (point O in
Figure 7.7a) and is
3.05
3.05
3.05
(24.81) a3.05 1
b 1 (38.7) a
b 1 (11.12 1 45.63) a
b
3
2
3
z5
5 1.81 m ■
120.26
7.4
A Generalized Case for Rankine Active Pressure
In Section 7.3, the relationship was developed for Rankine active pressure for a retaining
wall with a vertical back and a horizontal backfill. That can be extended to general cases
of frictionless walls with inclined backs and inclined backfills. Some of these cases will be
discussed in this section.
Granular Backfill
Figure 7.8 shows a retaining wall whose back is inclined at an angle u with the vertical.
The granular backfill is inclined at an angle a with the horizontal.
For a Rankine active case, the lateral earth pressure (sar ) at a depth z can be given
as (Chu, 1991),
sar 5
where ca 5 sin21 ¢
g z cos a"1 1 sin2 fr 2 2 sin fr cos ca
cos a 1 "sin2 fr 2 sin2 a
sin a
≤ 2 a 1 2u.
sin fr
(7.13)
(7.14)
z
a
H
Frictionless
wall
Figure 7.8 General case for a retaining
wall with granular backfill
7.4 A Generalized Case for Rankine Active Pressure
335
The pressure sar will be inclined at an angle br with the plane drawn at right angle
to the backface of the wall, and
br 5 tan21 ¢
sin fr sin ca
≤
1 2 sin fr cos ca
(7.15)
The active force Pa for unit length of the wall then can be calculated as
Pa 5
1
gH 2Ka
2
(7.16)
where
Ka 5
cos(a 2 u)"1 1 sin2 fr 2 2 sin fr cos ca
cos2 u Q cos a 1 "sin2 fr 2 sin2 a R
5 Rankine active earth-pressure coefficient for generalized case
(7.17)
The location and direction of the resultant force Pa is shown in Figure 7.9. Also shown
in this figure is the failure wedge, ABC. Note that BC will be inclined at an angle h. Or
h5
fr
p
a
1
sin a
1
1 2 sin21 ¢
≤
4
2
2
2
sin fr
(7.18)
C
A
Failure
wedge
Pa
H
H/3
B
4
1
sin sin1
2
2 2
sin (
)
Figure 7.9 Location and direction of
Rankine active force
336 Chapter 7: Lateral Earth Pressure
Granular Backfill with Vertical Back Face
As a special case, for a vertical backface of a wall (that is, u 5 0), as shown in Figure 7.10,
Eqs. (7.13), (7.16) and (7.17) simplify to the following.
If the backfill of a frictionless retaining wall is a granular soil (cr 5 0) and rises at an
angle a with respect to the horizontal (see Figure 7.10), the active earth-pressure coefficient
may be expressed in the form
Ka 5 cos a
cos a2"cos2 a2cos2 fr
cos a 1 "cos2 a2cos2 fr
(7.19)
where fr 5 angle of friction of soil.
At any depth z, the Rankine active pressure may be expressed as
sar 5 gzKa
(7.20)
Also, the total force per unit length of the wall is
Pa 5 12 gH 2Ka
(7.21)
Note that, in this case, the direction of the resultant force Pa is inclined at an angle a
with the horizontal and intersects the wall at a distance H>3 from the base of the
wall. Table 7.1 presents the values of Ka (active earth pressure) for various values of a
and fr.
a
z
Pa
H
H/3
Figure 7.10 Notations for active
pressure—Eqs. (7.19), (7.20), (7.21)
337
28
0.3610
0.3612
0.3618
0.3627
0.3639
0.3656
0.3676
0.3701
0.3730
0.3764
0.3802
0.3846
0.3896
0.3952
0.4015
0.4086
0.4165
0.4255
0.4357
0.4473
0.4605
0.4758
0.4936
0.5147
0.5404
0.5727
T
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
␣ (deg)
0.3470
0.3471
0.3476
0.3485
0.3496
0.3512
0.3531
0.3553
0.3580
0.3611
0.3646
0.3686
0.3731
0.3782
0.3839
0.3903
0.3975
0.4056
0.4146
0.4249
0.4365
0.4498
0.4651
0.4829
0.5041
0.5299
29
30
0.3333
0.3335
0.3339
0.3347
0.3358
0.3372
0.3389
0.3410
0.3435
0.3463
0.3495
0.3532
0.3573
0.3620
0.3671
0.3729
0.3794
0.3867
0.3948
0.4039
0.4142
0.4259
0.4392
0.4545
0.4724
0.4936
Table 7.1 Values of Ka [Eq. (7.19)]
0.3201
0.3202
0.3207
0.3214
0.3224
0.3237
0.3253
0.3272
0.3294
0.3320
0.3350
0.3383
0.3421
0.3464
0.3511
0.3564
0.3622
0.3688
0.3761
0.3842
0.3934
0.4037
0.4154
0.4287
0.4440
0.4619
31
0.3073
0.3074
0.3078
0.3084
0.3094
0.3105
0.3120
0.3138
0.3159
0.3182
0.3210
0.3241
0.3275
0.3314
0.3357
0.3405
0.3458
0.3518
0.3584
0.3657
0.3739
0.3830
0.3934
0.4050
0.4183
0.4336
32
0.2948
0.2949
0.2953
0.2959
0.2967
0.2978
0.2992
0.3008
0.3027
0.3049
0.3074
0.3103
0.3134
0.3170
0.3209
0.3253
0.3302
0.3356
0.3415
0.3481
0.3555
0.3637
0.3729
0.3832
0.3948
0.4081
33
0.2827
0.2828
0.2832
0.2837
0.2845
0.2855
0.2868
0.2883
0.2900
0.2921
0.2944
0.2970
0.2999
0.3031
0.3068
0.3108
0.3152
0.3201
0.3255
0.3315
0.3381
0.3455
0.3537
0.3628
0.3731
0.3847
34
␾ⴕ (deg) S
0.2710
0.2711
0.2714
0.2719
0.2726
0.2736
0.2747
0.2761
0.2778
0.2796
0.2818
0.2841
0.2868
0.2898
0.2931
0.2968
0.3008
0.3053
0.3102
0.3156
0.3216
0.3283
0.3356
0.3438
0.3529
0.3631
35
0.2596
0.2597
0.2600
0.2605
0.2611
0.2620
0.2631
0.2644
0.2659
0.2676
0.2696
0.2718
0.2742
0.2770
0.2800
0.2834
0.2871
0.2911
0.2956
0.3006
0.3060
0.3120
0.3186
0.3259
0.3341
0.3431
36
0.2486
0.2487
0.2489
0.2494
0.2500
0.2508
0.2518
0.2530
0.2544
0.2560
0.2578
0.2598
0.2621
0.2646
0.2674
0.2705
0.2739
0.2776
0.2817
0.2862
0.2911
0.2965
0.3025
0.3091
0.3164
0.3245
37
0.2379
0.2380
0.2382
0.2386
0.2392
0.2399
0.2409
0.2420
0.2432
0.2447
0.2464
0.2482
0.2503
0.2527
0.2552
0.2581
0.2612
0.2646
0.2683
0.2724
0.2769
0.2818
0.2872
0.2932
0.2997
0.3070
38
0.2275
0.2276
0.2278
0.2282
0.2287
0.2294
0.2303
0.2313
0.2325
0.2338
0.2354
0.2371
0.2390
0.2412
0.2435
0.2461
0.2490
0.2521
0.2555
0.2593
0.2634
0.2678
0.2727
0.2781
0.2840
0.2905
39
0.2174
0.2175
0.2177
0.2181
0.2186
0.2192
0.2200
0.2209
0.2220
0.2233
0.2247
0.2263
0.2281
0.2301
0.2322
0.2346
0.2373
0.2401
0.2433
0.2467
0.2504
0.2545
0.2590
0.2638
0.2692
0.2750
40
338 Chapter 7: Lateral Earth Pressure
Vertical Backface with c9– f9 Soil Backfill
For a retaining wall with a vertical back (u 5 0) and inclined backfill of cr–fr soil
(Mazindrani and Ganjali, 1997),
sar 5 gzKa 5 gzKar cos a
(7.22)
where
Kar 5
1
d
cos2 fr
2 cos2 a 1 2¢
cr
≤ cos fr sin fr
gz
cr 2
cr
2 B4 cos a(cos a2cos fr) 1 4¢ ≤ cos2 fr 1 8¢ ≤ cos2 a sin fr cos fr R
gz
gz
Å
2
2
t 21
2
(7.23)
Some values of Kar are given in Table 7.2. For a problem of this type, the depth of tensile
crack is given as
zc 5
2cr 1 1 sin fr
g Å 1 2 sin fr
(7.24)
For this case, the active pressure is inclined at an angle a with the horizontal.
Table 7.2 Values of Kar
c9
gz
f9 (deg)
a (deg)
0.025
0.05
0.1
0.5
15
0
5
10
15
0
5
10
15
0
5
10
15
0
5
10
15
0.550
0.566
0.621
0.776
0.455
0.465
0.497
0.567
0.374
0.381
0.402
0.443
0.305
0.309
0.323
0.350
0.512
0.525
0.571
0.683
0.420
0.429
0.456
0.514
0.342
0.348
0.366
0.401
0.276
0.280
0.292
0.315
0.435
0.445
0.477
0.546
0.350
0.357
0.377
0.417
0.278
0.283
0.296
0.321
0.218
0.221
0.230
0.246
20.179
20.184
20.186
20.196
20.210
20.212
20.218
20.229
20.231
20.233
20.239
20.250
20.244
20.246
20.252
20.263
20
25
30
7.4 A Generalized Case for Rankine Active Pressure
339
Example 7.4
Refer to the retaining wall in Figure 7.9. The backfill is granular soil. Given:
Wall:
H 5 10 ft
u 5 110°
Backfill:
a 5 15°
fr 5 35°
cr 5 0
g 5 110 lb>ft 3
Determine the Rankine active force, Pa, and its location and direction.
Solution
From Eq. (7.14),
ca 5 sin21 ¢
sin a
sin 15
≤ 2 a 1 2u 5 sin21 ¢
≤ 2 15 1 (2) (10) 5 31.82°
sin fr
sin 35
From Eq. (7.17),
Ka 5
5
cos(a 2 u)"1 1 sin2 fr 2 2 sin fr cos ca
cos2 u Q cos a 1 "sin2 fr 2 sin2 a R
cos(15 2 10)"1 1 sin2 35 2 (2) (sin 35) (sin 31.82)
cos2 10 Q cos 15 1 "sin2 35 2 sin2 15 R
5 0.59
Pa 5 1⁄2 gH 2Ka 5 ( 1⁄2 ) (110) (10) 2 (0.59) 5 3245 lb , ft
From Eq. (7.15),
br 5 tan21 ¢
sin fr sin ca
(sin 35) (sin 31.82)
≤ 5 tan21 B
R 5 30.58
1 2 sin fr cos ca
1 2 (sin 35) (cos 31.82)
The force Pa will act at a distance of 10>3 5 3.33 ft above the bottom of the wall and will
be inclined at an angle of130.5° to the normal drawn to the back face of the wall.
■
Example 7.5
For the retaining wall shown in Figure 7.10, H ⫽ 7.5 m, ␥ ⫽ 18 kN/m3, ␾⬘ ⫽ 20°,
c⬘ ⫽ 13.5 kN/m2, and ␣ ⫽ 10°. Calculate the Rankine active force, Pa, per unit length of
the wall and the location of the resultant force after the occurrence of the tensile crack.
Solution
From Eq. (7.24).
zr 5
(2) (13.5) 1 1 sin 20
2c r 1 1 sinfr
5
5 2.14 m
g Å 1 2 sinfr
18
Å 1 2 sin 20
340 Chapter 7: Lateral Earth Pressure
At z ⫽ 7.5 m,
cr
13.5
5
5 0.1
gz
(18) (7.5)
From Table 7.2, for ␾⬘ ⫽ 20°, c⬘/␥z ⫽ 0.1, and ␣ ⫽ 10°, the value of Ka⬘ is 0.377, so at
z ⫽ 7.5 m,
␴a⬘ ⫽ ␥zKa⬘cos ␣ ⫽ (18)(7.5)(0.377)(cos 10) ⫽ 50.1 kN/m2
After the occurrence of the tensile crack, the pressure distribution on the wall will be as
shown in Figure 7.11, so
1
Pa 5 a b (50.1) (7.5 2 2.14) 5 134.3 kN>m
2
and
z5
7.5 2 2.14
5 1.79 m
3
■
10°
2.14 m
5.36 m
Pa
z = 1.79 m
10°
7.5
Figure 7.11 Calculation of Rankine active force,
c⬘ ⫺ ␾⬘ soil
Coulomb’s Active Earth Pressure
The Rankine active earth pressure calculations discussed in the preceding sections were
based on the assumption that the wall is frictionless. In 1776, Coulomb proposed a theory
for calculating the lateral earth pressure on a retaining wall with granular soil backfill. This
theory takes wall friction into consideration.
To apply Coulomb’s active earth pressure theory, let us consider a retaining
wall with its back face inclined at an angle b with the horizontal, as shown in
Figure 7.12a. The backfill is a granular soil that slopes at an angle a with the horizontal.
7.5 Coulomb’s Active Earth Pressure
341
Pa(max)
Active
force
C3
C2
Wall movement
C1
away from
soil
A
Pa
c = 0
W
N
H
Pa
R
1 R
S
H/3
W
(b)
1
B
(a)
Figure 7.12 Coulomb’s active pressure
Also, let dr be the angle of friction between the soil and the wall (i.e., the angle of wall
friction).
Under active pressure, the wall will move away from the soil mass (to the left in the
figure). Coulomb assumed that, in such a case, the failure surface in the soil mass would
be a plane (e.g., BC1 , BC2 , c ). So, to find the active force, consider a possible soil failure wedge ABC1 . The forces acting on this wedge (per unit length at right angles to the
cross section shown) are as follows:
1. The weight of the wedge, W.
2. The resultant, R, of the normal and resisting shear forces along the surface, BC1 .
The force R will be inclined at an angle fr to the normal drawn to BC1 .
3. The active force per unit length of the wall, Pa , which will be inclined at an angle dr
to the normal drawn to the back face of the wall.
For equilibrium purposes, a force triangle can be drawn, as shown in Figure 7.12b.
Note that u1 is the angle that BC1 makes with the horizontal. Because the magnitude of W,
as well as the directions of all three forces, are known, the value of Pa can now be determined. Similarly, the active forces of other trial wedges, such as ABC2 , ABC3 , c, can
be determined. The maximum value of Pa thus determined is Coulomb’s active force (see
top part of Figure 7.12), which may be expressed as
Pa 5 12KagH 2
(7.25)
342 Chapter 7: Lateral Earth Pressure
where
Ka 5 Coulomb’s active earth pressure coefficient
5
sin2 (b 1 fr)
(7.26)
sin (fr 1 dr)sin (fr2a) 2
sin b sin (b2dr) B1 1
R
Å sin (b2dr)sin (a 1 b)
2
and H 5 height of the wall.
The values of the active earth pressure coefficient, Ka , for a vertical retaining
wall (b 5 90°) with horizontal backfill (a 5 0°) are given in Table 7.3. Note that
the line of action of the resultant force (Pa ) will act at a distance H>3 above the
base of the wall and will be inclined at an angle dr to the normal drawn to the back of
the wall.
In the actual design of retaining walls, the value of the wall friction angle dr is
assumed to be between fr>2 and 23fr. The active earth pressure coefficients for various values of fr, a, and b with dr 5 12fr and 32fr are respectively given in Tables 7.4 and 7.5.
These coefficients are very useful design considerations.
If a uniform surcharge of intensity q is located above the backfill, as shown in
Figure 7.13, the active force, Pa, can be calculated as
Pa 5 12KageqH 2
c
Eq. (7.25)
(7.27)
where
geq 5 g 1 c
sinb
2q
da b
sin (b 1 a) H
(7.28)
Table 7.3 Values of Ka [Eq. (7.26)] for b 5 90° and a 5 0°
d9 (deg)
f9 (deg)
0
5
10
15
20
25
28
30
32
34
36
38
40
42
0.3610
0.3333
0.3073
0.2827
0.2596
0.2379
0.2174
0.1982
0.3448
0.3189
0.2945
0.2714
0.2497
0.2292
0.2098
0.1916
0.3330
0.3085
0.2853
0.2633
0.2426
0.2230
0.2045
0.1870
0.3251
0.3014
0.2791
0.2579
0.2379
0.2190
0.2011
0.1841
0.3203
0.2973
0.2755
0.2549
0.2354
0.2169
0.1994
0.1828
0.3186
0.2956
0.2745
0.2542
0.2350
0.2167
0.1995
0.1831
7.5 Coulomb’s Active Earth Pressure
343
Table 7.4 Values of Ka [from Eq. (7.26)] for dr 5 23 fr
b (deg)
a (deg)
f9 (deg)
90
85
80
75
70
65
0
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
28
0.3213
0.3091
0.2973
0.2860
0.2750
0.2645
0.2543
0.2444
0.2349
0.2257
0.2168
0.2082
0.1998
0.1918
0.1840
0.3431
0.3295
0.3165
0.3039
0.2919
0.2803
0.2691
0.2583
0.2479
0.2379
0.2282
0.2188
0.2098
0.2011
0.1927
0.3702
0.3548
0.3400
0.3259
0.3123
0.2993
0.2868
0.2748
0.2633
0.2522
0.2415
0.2313
0.2214
0.2119
0.2027
0.4065
0.3588
0.3467
0.3349
0.3235
0.3125
0.3019
0.2916
0.2816
0.2719
0.2626
0.2535
0.2447
0.2361
0.2278
0.2197
0.3845
0.3709
0.3578
0.3451
0.3329
0.3211
0.3097
0.2987
0.2881
0.2778
0.2679
0.2582
0.2489
0.2398
0.2311
0.4164
0.4007
0.3857
0.3713
0.3575
0.3442
0.3314
0.3190
0.3072
0.2957
0.2846
0.2740
0.2636
0.2537
0.2441
0.4585
0.4007
0.3886
0.3769
0.3655
0.3545
0.3439
0.3335
0.3235
0.3137
0.3042
0.2950
0.2861
0.2774
0.2689
0.2606
0.4311
0.4175
0.4043
0.3916
0.3792
0.3673
0.3558
0.3446
0.3338
0.3233
0.3131
0.3033
0.2937
0.2844
0.2753
0.4686
0.4528
0.4376
0.4230
0.4089
0.3953
0.3822
0.3696
0.3574
0.3456
0.3342
0.3231
0.3125
0.3021
0.2921
0.5179
0.4481
0.4362
0.4245
0.4133
0.4023
0.3917
0.3813
0.3713
0.3615
0.3520
0.3427
0.3337
0.3249
0.3164
0.3080
0.4843
0.4707
0.4575
0.4447
0.4324
0.4204
0.4088
0.3975
0.3866
0.3759
0.3656
0.3556
0.3458
0.3363
0.3271
0.5287
0.5128
0.4974
0.4826
0.4683
0.4545
0.4412
0.4283
0.4158
0.4037
0.3920
0.3807
0.3697
0.3590
0.3487
0.5868
0.5026
0.4908
0.4794
0.4682
0.4574
0.4469
0.4367
0.4267
0.4170
0.4075
0.3983
0.3894
0.3806
0.3721
0.3637
0.5461
0.5325
0.5194
0.5067
0.4943
0.4823
0.4707
0.4594
0.4484
0.4377
0.4273
0.4172
0.4074
0.3978
0.3884
0.5992
0.5831
0.5676
0.5526
0.5382
0.5242
0.5107
0.4976
0.4849
0.4726
0.4607
0.4491
0.4379
0.4270
0.4164
0.6685
0.5662
0.5547
0.5435
0.5326
0.5220
0.5117
0.5017
0.4919
0.4824
0.4732
0.4641
0.4553
0.4468
0.4384
0.4302
0.6190
0.6056
0.5926
0.5800
0.5677
0.5558
0.5443
0.5330
0.5221
0.5115
0.5012
0.4911
0.4813
0.4718
0.4625
0.6834
0.6672
0.6516
0.6365
0.6219
0.6078
0.5942
0.5810
0.5682
0.5558
0.5437
0.5321
0.5207
0.5097
0.4990
0.7670
5
10
15
(continued)
344 Chapter 7: Lateral Earth Pressure
Table 7.4 (Continued)
b (deg)
a (deg)
f9 (deg)
90
85
80
75
70
65
20
29
30
31
32
33
34
35
36
37
38
39
40
41
42
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
0.3881
0.3707
0.3541
0.3384
0.3234
0.3091
0.2954
0.2823
0.2698
0.2578
0.2463
0.2353
0.2247
0.2146
0.4602
0.4364
0.4142
0.3935
0.3742
0.3559
0.3388
0.3225
0.3071
0.2925
0.2787
0.2654
0.2529
0.2408
0.2294
0.4397
0.4219
0.4049
0.3887
0.3732
0.3583
0.3442
0.3306
0.3175
0.3050
0.2929
0.2813
0.2702
0.2594
0.5205
0.4958
0.4728
0.4513
0.4311
0.4121
0.3941
0.3771
0.3609
0.3455
0.3308
0.3168
0.3034
0.2906
0.2784
0.4987
0.4804
0.4629
0.4462
0.4303
0.4150
0.4003
0.3862
0.3726
0.3595
0.3470
0.3348
0.3231
0.3118
0.5900
0.5642
0.5403
0.5179
0.4968
0.4769
0.4581
0.4402
0.4233
0.4071
0.3916
0.3768
0.3626
0.3490
0.3360
0.5672
0.5484
0.5305
0.5133
0.4969
0.4811
0.4659
0.4513
0.4373
0.4237
0.4106
0.3980
0.3858
0.3740
0.6714
0.6445
0.6195
0.5961
0.5741
0.5532
0.5335
0.5148
0.4969
0.4799
0.4636
0.4480
0.4331
0.4187
0.4049
0.6483
0.6291
0.6106
0.5930
0.5761
0.5598
0.5442
0.5291
0.5146
0.5006
0.4871
0.4740
0.4613
0.4491
0.7689
0.7406
0.7144
0.6898
0.6666
0.6448
0.6241
0.6044
0.5856
0.5677
0.5506
0.5342
0.5185
0.5033
0.4888
0.7463
0.7265
0.7076
0.6895
0.6721
0.6554
0.6393
0.6238
0.6089
0.5945
0.5805
0.5671
0.5541
0.5415
0.8880
0.8581
0.8303
0.8043
0.7799
0.7569
0.7351
0.7144
0.6947
0.6759
0.6579
0.6407
0.6242
0.6083
0.5930
Table 7.5 Values of Ka [from Eq. (7.26)] for dr 5 fr>2
b (deg)
a (deg)
f9 (deg)
90
85
80
75
70
65
0
28
29
30
31
32
33
34
35
36
0.3264
0.3137
0.3014
0.2896
0.2782
0.2671
0.2564
0.2461
0.2362
0.3629
0.3502
0.3379
0.3260
0.3145
0.3033
0.2925
0.2820
0.2718
0.4034
0.3907
0.3784
0.3665
0.3549
0.3436
0.3327
0.3221
0.3118
0.4490
0.4363
0.4241
0.4121
0.4005
0.3892
0.3782
0.3675
0.3571
0.5011
0.4886
0.4764
0.4645
0.4529
0.4415
0.4305
0.4197
0.4092
0.5616
0.5492
0.5371
0.5253
0.5137
0.5025
0.4915
0.4807
0.4702
7.5 Coulomb’s Active Earth Pressure
345
Table 7.5 (Continued)
b (deg)
a (deg)
5
10
15
f9 (deg)
90
85
80
75
70
65
37
38
39
40
41
42
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
28
29
30
31
32
33
34
35
0.2265
0.2172
0.2081
0.1994
0.1909
0.1828
0.3477
0.3337
0.3202
0.3072
0.2946
0.2825
0.2709
0.2596
0.2488
0.2383
0.2282
0.2185
0.2090
0.1999
0.1911
0.3743
0.3584
0.3432
0.3286
0.3145
0.3011
0.2881
0.2757
0.2637
0.2522
0.2412
0.2305
0.2202
0.2103
0.2007
0.4095
0.3908
0.3730
0.3560
0.3398
0.3244
0.3097
0.2956
0.2620
0.2524
0.2431
0.2341
0.2253
0.2168
0.3879
0.3737
0.3601
0.3470
0.3342
0.3219
0.3101
0.2986
0.2874
0.2767
0.2662
0.2561
0.2463
0.2368
0.2276
0.4187
0.4026
0.3872
0.3723
0.3580
0.3442
0.3309
0.3181
0.3058
0.2938
0.2823
0.2712
0.2604
0.2500
0.2400
0.4594
0.4402
0.4220
0.4046
0.3880
0.3721
0.3568
0.3422
0.3017
0.2920
0.2825
0.2732
0.2642
0.2554
0.4327
0.4185
0.4048
0.3915
0.3787
0.3662
0.3541
0.3424
0.3310
0.3199
0.3092
0.2988
0.2887
0.2788
0.2693
0.4688
0.4525
0.4368
0.4217
0.4071
0.3930
0.3793
0.3662
0.3534
0.3411
0.3292
0.3176
0.3064
0.2956
0.2850
0.5159
0.4964
0.4777
0.4598
0.4427
0.4262
0.4105
0.3953
0.3469
0.3370
0.3273
0.3179
0.3087
0.2997
0.4837
0.4694
0.4556
0.4422
0.4292
0.4166
0.4043
0.3924
0.3808
0.3695
0.3585
0.3478
0.3374
0.3273
0.3174
0.5261
0.5096
0.4936
0.4782
0.4633
0.4489
0.4350
0.4215
0.4084
0.3957
0.3833
0.3714
0.3597
0.3484
0.3375
0.5812
0.5611
0.5419
0.5235
0.5059
0.4889
0.4726
0.4569
0.3990
0.3890
0.3792
0.3696
0.3602
0.3511
0.5425
0.5282
0.5144
0.5009
0.4878
0.4750
0.4626
0.4505
0.4387
0.4272
0.4160
0.4050
0.3944
0.3840
0.3738
0.5928
0.5761
0.5599
0.5442
0.5290
0.5143
0.5000
0.4862
0.4727
0.4597
0.4470
0.4346
0.4226
0.4109
0.3995
0.6579
0.6373
0.6175
0.5985
0.5803
0.5627
0.5458
0.5295
0.4599
0.4498
0.4400
0.4304
0.4209
0.4177
0.6115
0.5972
0.5833
0.5698
0.5566
0.5437
0.5312
0.5190
0.5070
0.4954
0.4840
0.4729
0.4620
0.4514
0.4410
0.6719
0.6549
0.6385
0.6225
0.6071
0.5920
0.5775
0.5633
0.5495
0.5361
0.5230
0.5103
0.4979
0.4858
0.4740
0.7498
0.7284
0.7080
0.6884
0.6695
0.6513
0.6338
0.6168
(continued)
346 Chapter 7: Lateral Earth Pressure
Table 7.5 (Continued)
b (deg)
a (deg)
20
f9 (deg)
90
85
80
75
70
65
36
37
38
39
40
41
42
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
0.2821
0.2692
0.2569
0.2450
0.2336
0.2227
0.2122
0.4614
0.4374
0.4150
0.3941
0.3744
0.3559
0.3384
0.3218
0.3061
0.2911
0.2769
0.2633
0.2504
0.2381
0.2263
0.3282
0.3147
0.3017
0.2893
0.2773
0.2657
0.2546
0.5188
0.4940
0.4708
0.4491
0.4286
0.4093
0.3910
0.3736
0.3571
0.3413
0.3263
0.3120
0.2982
0.2851
0.2725
0.3807
0.3667
0.3531
0.3401
0.3275
0.3153
0.3035
0.5844
0.5586
0.5345
0.5119
0.4906
0.4704
0.4513
0.4331
0.4157
0.3991
0.3833
0.3681
0.3535
0.3395
0.3261
0.4417
0.4271
0.4130
0.3993
0.3861
0.3733
0.3609
0.6608
0.6339
0.6087
0.5851
0.5628
0.5417
0.5216
0.5025
0.4842
0.4668
0.4500
0.4340
0.4185
0.4037
0.3894
0.5138
0.4985
0.4838
0.4695
0.4557
0.4423
0.4293
0.7514
0.7232
0.6968
0.6720
0.6486
0.6264
0.6052
0.5851
0.5658
0.5474
0.5297
0.5127
0.4963
0.4805
0.4653
0.6004
0.5846
0.5692
0.5543
0.5399
0.5258
0.5122
0.8613
0.8313
0.8034
0.7772
0.7524
0.7289
0.7066
0.6853
0.6649
0.6453
0.6266
0.6085
0.5912
0.5744
0.5582
Surcharge = q
C
A
c = 0
H
Pa
B
KaH sin
sin ( + )
(a)
(b)
Figure 7.13 Coulomb’s active pressure with a surcharge on the
backfill
7.5 Coulomb’s Active Earth Pressure
347
Example 7.6
Consider the retaining wall shown in Figure 7.12a. Given: H 5 4.6 m; unit weight of
soil 5 16.5 kN>m3; angle of friction of soil 5 308; wall friction-angle, dr 5 23fr, soil
cohesion, c9 5 0; a 5 0, and b 5 908. Calculate the Coulomb’s active force per unit
length of the wall.
Solution
From Eq. (7.25)
Pa 5 12gH 2Ka
From Table 7.4, for a 5 08, b 5 908, f9 5 308, and d9 5 23fr 5208, Ka 5 0.297. Hence,
Pa 5 12 (16.5) (4.6) 2 (0.297) 5 51.85 kN , m
■
Example 7.7
Refer to Figure 7.13a. Given: H ⫽ 6.1 m, ␾⬘ ⫽ 30°, ␦⬘ ⫽ 20°, ␣ ⫽ 5°, ␤ ⫽ 85°,
q ⫽ 96 kN/m2, and ␥ ⫽ 18 kN/m3. Determine Coulomb’s active force and the location
of the line of action of the resultant Pa.
Solution
For ␤ ⫽ 85°, ␣ ⫽ 5°, ␦⬘ ⫽ 20°, ␾⬘ ⫽ 30°, and Ka ⫽ 0.3578 (Table 7.4). From
Eqs. (7.27) and (7.28),
2q
sinb
1
1
1
pa 5 KageqH 2 5 Ka cg 1
d H 2 5 KagH 2
2
2
H sin (b 1 a)
2
(')'*
sinb
1 KaHq c
d
Pa(1)
sin (b 1 a)
('''')''''*
Pa(2)
5 (0.5) (0.3578) (18) (6.1) 2 1 (0.3578) (6.1) (96) c
sin85
d
sin (85 1 5)
5 119.8 1 208.7 5 328.5 kN>m
Location of the line of action of the resultant:
Paz 5 Pa(1)
H
H
1 Pa(2)
3
2
or
(119.8) a
6.1
6.1
b 1 (208.7) a b
3
2
z5
328.5
5 2.68 m (measured vertically from the bottom of the wall)
■
348 Chapter 7: Lateral Earth Pressure
7.6
Lateral Earth Pressure Due to Surcharge
In several instances, the theory of elasticity is used to determine the lateral earth pressure
on unyielding retaining structures caused by various types of surcharge loading, such as
line loading (Figure 7.14a) and strip loading (Figure 7.14b).
According to the theory of elasticity, the stress at any depth, z, on a retaining structure caused by a line load of intensity q/unit length (Figure 7.14a) may be given as
s5
2q
a 2b
pH (a2 1 b2 ) 2
(7.29)
where ␴ ⫽ horizontal stress at depth z ⫽ bH
(See Figure 7.14a for explanations of the terms a and b.)
Line load
q / unit length
aH
z = bH
H
(a)
b
z
H
a
q / unit length
P
z
(b)
Figure 7.14 Lateral earth pressure caused by
(a) line load and (b) strip load
7.6 Lateral Earth Pressure Due to Surcharge
349
However, because soil is not a perfectly elastic medium, some deviations from Eq.
(7.29) may be expected. The modified forms of this equation generally accepted for use
with soils are as follows:
s5
4a
a2b
pH (a2 1 b2 )
for a . 0.4
(7.30)
s5
q
0.203b
H (0.16 1 b2 ) 2
for a # 0.4
(7.31)
and
Figure 7.14b shows a strip load with an intensity of q/unit area located at a distance
b⬘ from a wall of height H. Based on the theory of elasticity, the horizontal stress, ␴, at any
depth z on a retaining structure is
s5
q
(b 2 sin b cos 2a)
p
(7.32)
(The angles ␣ and ␤ are defined in Figure 7.14b.)
However, in the case of soils, the right-hand side of Eq. (7.32) is doubled to account
for the yielding soil continuum, or
s5
2q
(b 2 sinb cos 2a)
p
(7.33)
The total force per unit length (P) due to the strip loading only (Jarquio, 1981) may
be expressed as
P5
q
3H(u2 2 u1 )4
90
(7.34)
where
u1 5 tan2 a
b9
b
H
u2 5 tan21 a
(deg)
a 9 1 b9
b (deg)
H
(7.35)
(7.36)
The location z– (see Figure 7.14b) of the resultant force, P, can be given as
z5H2 c
H 2 (u2 2 u1 ) 1 (R 2 Q) 2 57.3arH
d
2H(u2 2 u1 )
(7.37)
350 Chapter 7: Lateral Earth Pressure
where
R ⫽ (a⬘ ⫹ b⬘)2(90 ⫺ ␪2)
(7.38)
Q ⫽ b⬘ (90 ⫺ ␪1)
(7.39)
2
Example 7.8
Refer to Figure 7.14b. Here, a⬘ ⫽ 2 m, b⬘ ⫽ 1 m, q ⫽ 40 kN/m2, and H ⫽ 6 m. Determine the total force on the wall (kN/m) caused by the strip loading only.
Solution From Eqs. (7.35) and (7.38),
1
u1 5 tan21 a b 5 9.46°
6
211
u2 5 tan21 a
b 5 26.57°
6
From Eq. (7.34)
P5
q
40
3H(u2 2 u1 )4 5
36(26.57 2 9.46)4 5 45.63 kN>m
90
90
■
Example 7.9
Refer to Example 7.8. Determine the location of the resultant –z .
Solution
From Eqs. (7.38) and (7.39),
R ⫽ (a⬘ ⫹ b⬘)2(90 ⫺ ␪2) ⫽ (2 ⫹ 1)2(90 ⫺ 26.57) ⫽ 570.87
Q ⫽ b⬘2(90 ⫺ ␪1) ⫽ (1)2(90 ⫺ 9.46) ⫽ 80.54
From Eq. (7.37),
z5H2 c
562 c
7.7
H 2 (u2 2 u1 ) 1 (R 2 Q) 2 57.3arH
d
2H(u2 2 u1 )
(6) 2 (26.57 2 9.46) 1 (570.87 2 80.54) 2 (57.3) (2) (6)
d 5 3.96 m
(2) (6) (26.57 2 9.46)
■
Active Earth Pressure for Earthquake Conditions
Coulomb’s active earth pressure theory (see Section 7.5) can be extended to take into account
the forces caused by an earthquake. Figure 7.15 shows a condition of active pressure with a
granular backfill (c9 5 0). Note that the forces acting on the soil failure wedge in Figure 7.15
7.7 Active Earth Pressure for Earthquake Conditions
351
kW
Granular backfill
c = 0
khW
W
H
Pae
Figure 7.15 Derivation of Eq. (7.42)
are essentially the same as those shown in Figure 7.12a with the addition of khW and kvW
in the horizontal and vertical direction respectively; kh and kv may be defined as
kh 5
horizontal earthquake acceleration component
acceleration due to gravity, g
(7.40)
kv 5
vertical earthquake acceleration component
acceleration due to gravity, g
(7.41)
As in Section 7.5, the relation for the active force per unit length of the wall (Pae) can
be determined as
Pae 5 12gH 2 (1 2 kv )Kae
(7.42)
where
Kae 5 active earth pressure coefficient
5
sin2 (fr 1 b 2 ur)
cos ur sin2 b sin(b 2 ur 2 dr) B1 1
sin (fr 1 dr)sin (fr 2 ur 2 a) 2
R
Å sin (b 2 dr 2 ur)sin (a 1 b)
(7.43)
ur 5 tan21 c
kh
d
(1 2 kv )
(7.44)
352 Chapter 7: Lateral Earth Pressure
Note that for no earthquake condition
kh 5 0,
kv 5 0,
and
ur 5 0
Hence Kae 5 Ka [as given by Eq. (7.26)]. Some values of Kae for b 5 908 and kv 5 0 are
given in Table 7.6.
The magnitude of Pae as given in Eq. (7.42) also can be determined as (Seed and
Whitman, 1970),
sin2b9
1
Pae 5 gH 2 (1 2 kv ) 3Ka (b9,a9 )4 a
b
2
cosu9sin2b
(7.45)
where
␤⬘ ⫽ ␤ ⫺ ␪⬘
(7.46)
␣⬘ ⫽ ␪⬘ ⫹ ␣
(7.47)
Ka(␤⬘,␣⬘) ⫽ Coulomb’s active earth-pressure coefficient on a wall with a back face inclination of ␤⬘ with the horizontal and with a back fill inclined at an angle ␣⬘
with the horizontal (such as Tables 7.4 and 7.5)
Equation (7.42) is usually referred to as the Mononobe–Okabe solution. Unlike the
case shown in Figure 7.12a, the resultant earth pressure in this situation, as calculated by
Eq. (7.42) does not act at a distance of H>3 from the bottom of the wall. The following procedure may be used to obtain the location of the resultant force Pae:
Step 1. Calculate Pae by using Eq. (7.42)
Step 2. Calculate Pa by using Eq. (7.25)
Step 3. Calculate
DPae 5 Pae 2 Pa
(7.48)
Step 4. Assume that Pa acts at a distance of H>3 from the bottom of the wall
(Figure 7.16)
Pae
Pa
H
0.6 H
H/3
Figure 7.16 Determining the line of action of Pae
7.7 Active Earth Pressure for Earthquake Conditions
353
Table 7.6 Values of Kae [Eq. (7.43)] for b 5 908 and kv 5 0
f9(deg)
353
kh
d9 (deg)
a(deg)
28
30
35
40
45
0.1
0.2
0.3
0.4
0.5
0
0
0.427
0.508
0.611
0.753
1.005
0.397
0.473
0.569
0.697
0.890
0.328
0.396
0.478
0.581
0.716
0.268
0.382
0.400
0.488
0.596
0.217
0.270
0.334
0.409
0.500
0.1
0.2
0.3
0.4
0.5
0
5
0.457
0.554
0.690
0.942
—
0.423
0.514
0.635
0.825
—
0.347
0.424
0.522
0.653
0.855
0.282
0.349
0.431
0.535
0.673
0.227
0.285
0.356
0.442
0.551
0.1
0.2
0.3
0.4
0.5
0
10
0.497
0.623
0.856
—
—
0.457
0.570
0.748
—
—
0.371
0.461
0.585
0.780
—
0.299
0.375
0.472
0.604
0.809
0.238
0.303
0.383
0.486
0.624
0.1
0.2
0.3
0.4
0.5
fr>2
0
0.396
0.485
0.604
0.778
1.115
0.368
0.452
0.563
0.718
0.972
0.306
0.380
0.474
0.599
0.774
0.253
0.319
0.402
0.508
0.648
0.207
0.267
0.340
0.433
0.522
0.1
0.2
0.3
0.4
0.5
fr>2
5
0.428
0.537
0.699
1.025
—
0.396
0.497
0.640
0.881
—
0.326
0.412
0.526
0.690
0.962
0.268
0.342
0.438
0.568
0.752
0.218
0.283
0.367
0.475
0.620
0.1
0.2
0.3
0.4
0.5
fr>2
10
0.472
0.616
0.908
—
—
0.433
0.562
0.780
—
—
0.352
0.454
0.602
0.857
—
0.285
0.371
0.487
0.656
0.944
0.230
0.303
0.400
0.531
0.722
0.1
0.2
0.3
0.4
0.5
2
3 fr
0
0.393
0.486
0.612
0.801
1.177
0.366
0.454
0.572
0.740
1.023
0.306
0.384
0.486
0.622
0.819
0.256
0.326
0.416
0.533
0.693
0.212
0.276
0.357
0.462
0.600
0.1
0.2
0.3
0.4
0.5
2
3 fr
5
0.427
0.541
0.714
1.073
—
0.395
0.501
0.655
0.921
—
0.327
0.418
0.541
0.722
1.034
0.271
0.350
0.455
0.600
0.812
0.224
0.294
0.386
0.509
0.679
0.1
0.2
0.3
0.4
0.5
2
3 fr
10
0.472
0.625
0.942
—
—
0.434
0.570
0.807
—
—
0.354
0.463
0.624
0.909
—
0.290
0.381
0.509
0.699
1.037
0.237
0.317
0.423
0.573
0.800
354 Chapter 7: Lateral Earth Pressure
Step 5. Assume that DPae acts at a distance of 0.6H from the bottom of the wall
(Figure 7.16)
Step 6. Calculate the location of the resultant as
z5
(0.6H) (DPae ) 1 a
H
b (Pa )
3
Pae
Example 7.10
Refer to Figure 7.17. For kv 5 0 and kh 5 0.3, determine:
a. Pae using Eq. (7.45)
b. The location of the resultant, z, from the bottom of the wall
35
16.51 kN/m3
17.5
3.05 m
Figure 7.17
Solution
Part a
From Eq. (7.44),
u9 5 tan21 a
kh
0.3
b 5 tan21 a
b 5 16.7°
1 2 kv
120
From Eqs. (7.46) and (7.47),
␤⬘ ⫽ ␤ ⫺ ␪⬘ ⫽ 90 ⫺ 16.7 ⫽ 73.3°
␣⬘ ⫽ ␪⬘ ⫹ ␣ ⫽ 16.7 ⫹ 0 ⫽ 16.7°
17.5
d9
5
5 0.5
9
35
f
(7.49)
7.8 Active Pressure for Wall Rotation about the Top: Braced Cut
355
We will refer to Table 7.5. For ␾⬘ ⫽ 35°, ␦⬘/␾⬘ ⫽ 0.5, ␤⬘ ⫽ 73.3°, and ␣⬘ ⫽ 16.7°, the
value of Ka(␤⬘,␣⬘) ⬇ 0.495. Thus, from Eq. (7.45),
sin2br
1
gH 2 (1 2 kv ) 3Ka (br,ar)4 a
b
2
cos ursin2b
1
sin273.3
5 (16.51) (3.05) 2 (1 2 0) (0.495) a
b 5 36.4 kN>m
2
cos16.7 sin290
Pae 5
Part b
From Eq. (7.25),
Pa 5 12gH 2Ka
From Eq. (7.26) with d9 5 17.58, b 5 908, and a 5 08, Ka < 0.246 (Table 7.5).
Pa 5 12 (16.51) (3.05) 2 (0.246) 5 18.89 kN>m
DPae 5 Pae 2 Pa 5 36.32 2 18.89 5 17.43 kN>m
From Eq. (7.49),
z5
5
7.8
(0.6H) ( DPae ) 1 (H>3) (Pa )
Pae
3(0.6) (3.05) 4 (17.43) 1 (3.05>3) (18.89)
36.32
5 1.41 m
■
Active Pressure for Wall Rotation about the Top:
Braced Cut
In the preceding sections, we have seen that a retaining wall rotates about its bottom.
(See Figure 7.18a.) With sufficient yielding of the wall, the lateral earth pressure is
approximately equal to that obtained by Rankine’s theory or Coulomb’s theory. In
contrast to retaining walls, braced cuts show a different type of wall yielding. (See
Figure 7.18b.) In this case, deformation of the wall gradually increases with the depth of
excavation. The variation of the amount of deformation depends on several factors, such
as the type of soil, the depth of excavation, and the workmanship involved. However,
with very little wall yielding at the top of the cut, the lateral earth pressure will be close
to the at-rest pressure. At the bottom of the wall, with a much larger degree of yielding,
the lateral earth pressure will be substantially lower than the Rankine active earth
pressure. As a result, the distribution of lateral earth pressure will vary substantially in
comparison to the linear distribution assumed in the case of retaining walls.
The total lateral force per unit length of the wall, Pa , imposed on a wall may be evaluated theoretically by using Terzaghi’s (1943) general wedge theory. (See
Figure 7.19.) The failure surface is assumed to be the arc of a logarithmic spiral, defined as
r 5 roeu tan fr
where fr 5 effective angle of friction of soil.
(7.50)
356 Chapter 7: Lateral Earth Pressure
Strut
(a)
(b)
Figure 7.18 Nature of yielding of walls: (a) retaining wall; (b) braced cut
Center of
log spiral
c
90
Lateral pressure, a
b
c
Ca
naH
Pa
W
H
C
a
R
Figure 7.19 Braced cut analysis
by general wedge theory: wall
rotation about top
Arc of
log spiral
In the figure, H is the height of the cut, and the unit weight, angle of friction, and
cohesion of the soil are equal to g, fr, and cr, respectively. Following are the forces per
unit length of the cut acting on the trial failure wedge:
1.
2.
3.
4.
5.
Weight of the wedge, W
Resultant of the normal and shear forces along ab, R
Cohesive force along ab, C
Adhesive force along ac, Ca
Pa , which is the force acting a distance naH from the bottom of the wall and is
inclined at an angle d9 to the horizontal
The adhesive force is
Ca 5 carH
(7.51)
where car 5 unit adhesion.
A detailed outline for the evaluation of Pa is beyond the scope of this text; those
interested should check a soil mechanics text for more information. Kim and Preber
(1969) provided tabulated values of Pa> 12gH 2 determined by using the principles of general wedge theory. Table 7.7 gives the variation of Pa>0.5gH 2 for granular soil backfill
obtained using the general wedge theory.
357
7.9 Active Earth Pressure for Translation of Retaining Wall—Granular Backfil
Table 7.7 Active Pressure for Wall Rotation—General Wedge Theory
(Granular Soil Backfill)
Soil friction
angle, f9(deg)
25
Pa , 0.5 gH 2
d9 , f9
na 5 0.3
na 5 0.4
na 5 0.5
na 5 0.6
0
0.371
0.345
0.342
0.344
0.405
0.376
0.373
0.375
0.447
0.413
0.410
0.413
0.499
0.460
0.457
0.461
0.304
0.282
0.281
0.289
0.330
0.306
0.305
0.313
0.361
0.334
0.332
0.341
0.400
0.386
0.367
0.377
0.247
0.231
0.232
0.243
0.267
0.249
0.249
0.262
0.290
0.269
0.270
0.289
0.318
0.295
0.296
0.312
0.198
0.187
0.190
0.197
0.213
0.200
0.204
0.211
0.230
0.216
0.220
0.228
0.252
0.235
0.239
0.248
0.205
0.149
0.153
0.173
0.220
0.159
0.164
0.184
0.237
0.171
0.176
0.198
0.259
0.185
0.196
0.215
1⁄2
2⁄3
1
30
0
1⁄2
2⁄3
1
35
0
1⁄2
2⁄3
1
40
0
1⁄2
2⁄3
1
45
0
1⁄2
2⁄3
1
7.9
Active Earth Pressure for Translation
of Retaining Wall—Granular Backfill
Under certain circumstances, retaining walls may undergo lateral translation, as shown in
Figure 7.20. A solution to the distribution of active pressure for this case was provided by
Dubrova (1963) and was also described by Harr (1966). The solution of Dubrova assumes
the validity of Coulomb’s solution [Eqs. (7.25) and (7.26)]. In order to understand this procedure, let us consider a vertical wall with a horizontal granular backfill (Figure 7.21). For
rotation about the top of the wall, the resultant R of the normal and shear forces along the
rupture line AC is inclined at an angle f9 to the normal drawn to AC. According to Dubrova
there exists infinite number of quasi-rupture lines such as A9C9, A0C0, . . . for which the
resultant force R is inclined at an angle c, where
c5
frz
H
(7.52)
Now, refer to Eqs. (7.25) and (7.26) for Coulomb’s active pressure. For b 5 908 and a 5 0,
the relationship for Coulomb’s active force can also be rewritten as
2
Pa 5
g
D
2 cos dr
H
T
1
1 (tan2 fr 1 tan fr tan dr) 0.5
cos fr
(7.53)
358 Chapter 7: Lateral Earth Pressure
B
C
z
A
R
H
C
C
R
Granular
backfill
c 0
A
R
A
Figure 7.20 Lateral
translation of retaining wall
Figure 7.21 Quasi-rupture lines behind a retaining wall
The force against the wall at any z is then given as
2
Pa 5
g
z
D
T
2 cos dr
1
2
0.5
1 (tan c 1 tan c tan dr)
cos c
(7.54)
The active pressure at any depth z for wall rotation about the top is
sar (z) 5
dPa
g
z cos2 c
z2fr cos2 c
<
c
2
(sin c 1 m) d (7.55)
2
dz
cos dr (1 1 m sin c)
H(1 1 m sin c)
where m 5 a1 1
tan dr 0.5
b .
tan c
(7.56)
For frictionless walls, d9 5 0 and Eq. (7.55) simplifies to
sar (z) 5 g tan2 a45 2
c
frz2
b az 2
b
2
H cos c
(7.57)
For wall rotation about the bottom, a similar expression can be found in the form
sar (z) 5
2
gz
cos fr
a
b
cos dr 1 1 m sin fr
(7.58)
For translation of the wall, the active pressure can then be taken as
sar (z) translation 5 12 3sar (z) rotation about top 1 sar (z) rotation about bottom4
(7.59)
7.9 Active Earth Pressure for Translation of Retaining Wall—Granular Backfil
359
Example 7.11
Consider a frictionless wall 5 m high. For the granular backfill, g 5 17.3 kN/m3 and f9 5
368. Calculate and plot the variation of sa(z) for a translation mode of the wall movement.
Solution
For a frictionless wall, d9 5 0. Hence, m is equal to one [Eq. (7.56)]. So for rotation
about the top, from Eq. (7.57),
sar (z) 5 sa(1)
r 5 g tan2 a45 2
frz
b Dz 2
2H
frz2
T
frz
b
H cos a
H
For rotation about the bottom, from Eq. (7.58),
sar (z) 5 sa(2)
r 5 gz a
sar (z) translation 5
2
cos fr
b
1 1 sin fr
sa(1)
r 1 sa(2)
r
2
The following table can now be prepared with g 5 17.3 kN/m3, f9 5 368, and H 5 5 m.
z
(m)
sa9(1)
(kN , m2)
sa9(2)
(kN , m2)
s9a(z) translation
(kN , m2)
0
1.25
2.5
3.75
5.0
0
13.26
15.26
11.48
5.02
0
5.62
11.24
16.86
22.48
0
9.44
13.25
14.17
13.75
The plot of sa(z) versus z is shown in Figure 7.22
a(z) translation (kN/m2)
0
5
10
15
0
z (m)
1.25
2.5
3.75
5
Figure 7.22
■
360 Chapter 7: Lateral Earth Pressure
Passive Pressure
7.10
Rankine Passive Earth Pressure
Figure 7.23a shows a vertical frictionless retaining wall with a horizontal backfill. At depth z,
the effective vertical pressure on a soil element is sor 5 gz. Initially, if the wall does not yield
at all, the lateral stress at that depth will be shr 5 Kosor. This state of stress is illustrated by
the Mohr’s circle a in Figure 7.23b. Now, if the wall is pushed into the soil mass by an amount
Dx, as shown in Figure 7.23a, the vertical stress at depth z will stay the same; however, the
horizontal stress will increase. Thus, shr will be greater than Kosor . The state of stress can now
be represented by the Mohr’s circle b in Figure 7.23b. If the wall moves farther inward (i.e.,
Dx is increased still more), the stresses at depth z will ultimately reach the state represented
by Mohr’s circle c. Note that this Mohr’s circle touches the Mohr–Coulomb failure envelope,
which implies that the soil behind the wall will fail by being pushed upward. The horizontal
stress, shr , at this point is referred to as the Rankine passive pressure, or shr 5 spr .
For Mohr’s circle c in Figure 7.23b, the major principal stress is spr , and the minor
principal stress is sor . Substituting these quantities into Eq. (1.87) yields
spr 5 sor tan2 ¢ 45 1
fr
fr
≤ 1 2crtan ¢45 1 ≤
2
2
(7.60)
Now, let
Kp 5 Rankine passive earth pressure coefficient
5 tan2 ¢45 1
fr
≤
2
(7.61)
Then, from Eq. (7.60), we have
spr 5 sor Kp 1 2cr"Kp
(7.62)
Equation (7.62) produces (Figure 7.23c), the passive pressure diagram for the wall
shown in Figure 7.23a. Note that at z 5 0,
sor 5 0
and spr 5 2cr"Kp
and at z 5 H,
sor 5 gH
and spr 5 gHKp 1 2cr"Kp
The passive force per unit length of the wall can be determined from the area of the
pressure diagram, or
Pp 5 12gH 2Kp 1 2crH"Kp
(7.63)
The approximate magnitudes of the wall movements, Dx, required to develop failure
under passive conditions are as follows:
7.10 Rankine Passive Earth Pressure
Wall movement for
passive condition, D x
Soil type
Dense sand
Loose sand
Stiff clay
Soft clay
0.005H
0.01H
0.01H
0.05H
Direction of
wall movement
x
o
z
45 /2
45 /2
z
c
H
h
Rotation about
this point
(a)
Shear stress
s c
c
n ta
c
a
Effective
normal
stress
h p
b
h Koo h
o
(b)
2c Kp
H
Kp H 2c Kp
(c)
Figure 7.23 Rankine passive pressure
361
362 Chapter 7: Lateral Earth Pressure
If the backfill behind the wall is a granular soil (i.e., cr 5 0), then, from Eq. (7.63), the
passive force per unit length of the wall will be
Pp 5
1
gH 2Kp
2
(7.64)
Example 7.12
A 3-m high wall is shown in Figure 7.24a. Determine the Rankine passive force per unit
length of the wall.
Solution
For the top layer
Kp(1) 5 tan2 a45 1
z
2m
f1r
b 5 tan2 (45 1 15) 5 3
2
15.72 kN/m3
1 30
c1 0
Groundwater
table
1
112.49
94.32
1m
sat 18.86 kN/m3
2 26
c2 10 kN/m2
2
3
0
(a)
135.65
kN/m2
(b)
4
Figure 7.24
From the bottom soil layer
Kp(2) 5 tan2 a45 1
f2r
b 5 tan2 (45 1 13) 5 2.56
2
spr 5 sor Kp 1 2cr"Kp
where
sor 5 effective vertical stress
at z 5 0, sor 5 0, c1r 5 0, spr 5 0
at z 5 2 m, sor 5 (15.72) (2) 5 31.44 kN>m2, c1r 5 0
So, for the top soil layer
spr 5 31.44Kp(1) 1 2(0)"Kp(1) 5 31.44(3) 5 94.32 kN>m2
9.81
kN/m2
7.11 Rankine Passive Earth Pressure: Vertical Backface and Inclined Backfill
363
At this depth, that is z 5 2 m, for the bottom soil layer
spr 5 sor Kp(2) 1 2c2r "Kp(2) 5 31.44(2.56) 1 2(10)"2.56
5 80.49 1 32 5 112.49 kN>m2
Again, at z 5 3 m,
sor 5 (15.72) (2) 1 (gsat 2 gw ) (1)
5 31.44 1 (18.86 2 9.81) (1) 5 40.49 kN>m2
Hence,
spr 5 sor Kp(2) 1 2c2r "Kp(2) 5 40.49(2.56) 1 (2) (10) (1.6)
5 135.65 kN , m2
Note that, because a water table is present, the hydrostatic stress, u, also has to be taken into
consideration. For z 5 0 to 2 m, u 5 0; z 5 3 m, u 5 (1)(gw) 5 9.81 kN>m2.
The passive pressure diagram is plotted in Figure 6.24b. The passive force per unit
length of the wall can be determined from the area of the pressure diagram as follows:
Area
no.
1
2
3
4
Area
(12 )
(2)(94.32)
(112.49)(1)
(12 ) (1)(135.65 2 112.49)
(12 ) (9.81)(1)
5 94.32
5 112.49
5 11.58
5 4.905
PP < 223.3 kN , m
7.11
■
Rankine Passive Earth Pressure: Vertical Backface
and Inclined Backfill
Granular Soil
For a frictionless vertical retaining wall (Figure 7.10) with a granular backfill (cr 5 0),
the Rankine passive pressure at any depth can be determined in a manner similar to that
done in the case of active pressure in Section 7.4. The pressure is
spr 5 gzKp
(7.65)
Pp 5 12gH 2Kp
(7.66)
and the passive force is
where
Kp 5 cos a
cos a 1 "cos2 a 2 cos2 fr
cos a2"cos2 a 2 cos2 fr
(7.67)
364 Chapter 7: Lateral Earth Pressure
Table 7.8 Passive Earth Pressure Coefficient Kp [from Eq. (7.67)]
S
f9 (deg)S
Ta (deg)
28
30
32
34
36
38
40
0
5
10
15
20
25
2.770
2.715
2.551
2.284
1.918
1.434
3.000
2.943
2.775
2.502
2.132
1.664
3.255
3.196
3.022
2.740
2.362
1.894
3.537
3.476
3.295
3.003
2.612
2.135
3.852
3.788
3.598
3.293
2.886
2.394
4.204
4.136
3.937
3.615
3.189
2.676
4.599
4.527
4.316
3.977
3.526
2.987
As in the case of the active force, the resultant force, Pp , is inclined at an angle a with the
horizontal and intersects the wall at a distance H>3 from the bottom of the wall. The values of
Kp (the passive earth pressure coefficient) for various values of a and fr are given in Table 7.8.
c 9–f9 Soil
If the backfill of the frictionless vertical retaining wall is a c9–f9 soil (see Figure 7.10), then
(Mazindrani and Ganjali, 1997)
sar 5 gzKp 5 gzKpr cos a
(7.68)
where
1
Kpr 5
d
cos2 fr
cr
≤cos fr sin fr
gz
t 21
cr 2 2
cr
2
2
2
2
1 4 cos a(cos a 2 cos fr) 1 4 ¢ ≤ cos fr 1 8¢ ≤cos a sin fr cos fr
gz
gz
Å
(7.69)
2 cos2 a 1 2 ¢
The variation of Kpr with fr, a, and cr>gz is given in Table 7.9 (Mazindrani and
Ganjali, 1997).
Table 7.9 Values of Kpr
c9 , gz
f9 (deg)
a (deg)
0.025
0.050
0.100
0.500
15
0
5
10
15
0
5
10
15
0
5
10
15
1.764
1.716
1.564
1.251
2.111
2.067
1.932
1.696
2.542
2.499
2.368
2.147
1.829
1.783
1.641
1.370
2.182
2.140
2.010
1.786
2.621
2.578
2.450
2.236
1.959
1.917
1.788
1.561
2.325
2.285
2.162
1.956
2.778
2.737
2.614
2.409
3.002
2.971
2.880
2.732
3.468
3.435
3.339
3.183
4.034
3.999
3.895
3.726
20
25
7.12 Coulomb’s Passive Earth Pressure
365
Table 7.9 (Continued)
c9 , gz
7.12
f9 (deg)
a (deg)
0.025
0.050
0.100
0.500
30
0
5
10
15
3.087
3.042
2.907
2.684
3.173
3.129
2.996
2.777
3.346
3.303
3.174
2.961
4.732
4.674
4.579
4.394
Coulomb’s Passive Earth Pressure
Coulomb (1776) also presented an analysis for determining the passive earth pressure (i.e.,
when the wall moves into the soil mass) for walls possessing friction (dr 5 angle of wall
friction) and retaining a granular backfill material similar to that discussed in Section 7.5.
To understand the determination of Coulomb’s passive force, Pp , consider the wall
shown in Figure 7.25a. As in the case of active pressure, Coulomb assumed that the potential
failure surface in soil is a plane. For a trial failure wedge of soil, such as ABC1 , the forces per
unit length of the wall acting on the wedge are
1. The weight of the wedge, W
2. The resultant, R, of the normal and shear forces on the plane BC1 , and
3. The passive force, Pp
Passive force
Pp(min)
Wall movement
toward
the soil
C2
C3
C1
A
Pp
R
H
W
Pp
H/3
1
c 0
R
W
1 B
(a)
Figure 7.25 Coulomb’s passive pressure
(b)
366 Chapter 7: Lateral Earth Pressure
Table 7.10 Values of Kp [from Eq. (7.71)] for b 5 90° and a 5 0°
d9 (deg)
f9 (deg)
0
5
10
15
20
15
20
25
30
35
40
1.698
2.040
2.464
3.000
3.690
4.600
1.900
2.313
2.830
3.506
4.390
5.590
2.130
2.636
3.286
4.143
5.310
6.946
2.405
3.030
3.855
4.977
6.854
8.870
2.735
3.525
4.597
6.105
8.324
11.772
Figure 7.25b shows the force triangle at equilibrium for the trial wedge ABC1 . From
this force triangle, the value of Pp can be determined, because the direction of all three forces
and the magnitude of one force are known.
Similar force triangles for several trial wedges, such as ABC1, ABC2, ABC3, c,
can be constructed, and the corresponding values of Pp can be determined. The top part of
Figure 7.25a shows the nature of variation of the Pp values for different wedges. The minimum value of Pp in this diagram is Coulomb’s passive force, mathematically expressed as
Pp 5 12 gH 2Kp
(7.70)
where
Kp 5 Coulomb’s passive pressure coefficient
5
sin2 (b2fr)
sin (fr 1 dr)sin (fr 1 a) 2
sin b sin (b 1 dr) B1 2
R
Å sin (b 1 dr)sin (b 1 a)
(7.71)
2
The values of the passive pressure coefficient, Kp , for various values of fr and dr are given
in Table 7.10 (b 5 90°, a 5 0°).
Note that the resultant passive force, Pp , will act at a distance H>3 from the bottom of
the wall and will be inclined at an angle d9 to the normal drawn to the back face of the wall.
7.13
Comments on the Failure Surface Assumption
for Coulomb’s Pressure Calculations
Coulomb’s pressure calculation methods for active and passive pressure have been discussed
in Sections 7.5 and 7.12. The fundamental assumption in these analyses is the acceptance of
plane failure surface. However, for walls with friction, this assumption does not hold in practice. The nature of actual failure surface in the soil mass for active and passive pressure is
shown in Figure 7.26a and b, respectively (for a vertical wall with a horizontal backfill). Note
that the failure surface BC is curved and that the failure surface CD is a plane.
Although the actual failure surface in soil for the case of active pressure is somewhat different from that assumed in the calculation of the Coulomb pressure, the results are not greatly
different. However, in the case of passive pressure, as the value of dr increases, Coulomb’s
367
7.13 Comments on the Failure Surface Assumption for Coulomb’s Pressure Calculations
A
D
45 /2
2
H
3
45 /2
H
Pa
C
B
(a)
C
A
D
45 /2
45 /2
2
H 3H
Pp
C
B
(b)
Figure 7.26 Nature of failure surface in soil with wall friction: (a) active pressure;
(b) passive pressure
method of calculation gives increasingly erroneous values of Pp . This factor of error could lead
to an unsafe condition because the values of Pp would become higher than the soil resistance.
Several studies have been conducted to determine the passive force Pp , assuming that
the curved portion BC in Figure 7.26b is an arc of a circle, an ellipse, or a logarithmic spiral.
Shields and Tolunay (1973) analyzed the problem of passive pressure for a vertical
wall with a horizontal granular soil backfill (cr 5 0). This analysis was done by considering the stability of the wedge ABCCr (see Figure 7.26b), using the method of slices
and assuming BC as an arc of a logarthmic spiral. From Figure 7.26b, the passive force
per unit length of the wall can be expressed as
1
Pp 5 KpgH 2
2
(7.72)
The values of the passive earth-pressure coefficient, Kp , obtained by Shields and Tolunay
are given in Figure 7.27. These are as good as any for design purposes.
Solution by the Method of Triangular Slices
Zhu and Qian (2000) used the method of triangular slices (such as in the zone of ABC in
Fig. 7.28) to obtain the variation of Kp. According to this analysis
Kp ⫽ Kp(␦⬘ ⫽ 0)R
(7.73)
where
Kp ⫽ passive earth-pressure coefficient for a given value of ␪, ␦⬘, and ␾⬘
Kp(␦⬘ ⫽ 0) ⫽ Kp for a given value of ␪, ␾⬘ with ␦⬘ ⫽ 0
R ⫽ modification factor which is a function of ␾⬘, ␪, ␦⬘/␾⬘
368 Chapter 7: Lateral Earth Pressure
18
16
1
14
0.8
12
0.6
10
Kp
0.4
8
0.2
6
0
4
2
0
20
25
30
35
Soil friction angle, (deg)
40
45
Figure 7.27 Kp based on Shields and Tolunay’s analysis
A
C′
45 – ′/2
45 – ′/2
Granular soil
′
H
′
H
3
C
B
Figure 7.28 Passive pressure solution by the method of triangular slices
(Note: BC is arc of a logarthimic spiral)
7.13 Comments on the Failure Surface Assumption for Coulomb’s Pressure Calculations
369
The variations of Kp(␦⬘ ⫽ 0) are given in Table 7.11. The interpolated values of R are
given in Table 7.12
Table 7.11 Variation of Kp(␦⬘ ⫽ 0) [see Eq. (7.73) and Figure 7.28]*
␪ (deg)
␾⬘(deg)
30
25
20
15
10
5
0
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
1.70
1.74
1.77
1.81
1.84
1.88
1.91
1.95
1.99
2.03
2.07
2.11
2.15
2.20
2.24
2.29
2.33
2.38
2.43
2.48
2.53
2.59
2.64
2.70
2.76
2.82
1.69
1.73
1.77
1.81
1.85
1.89
1.93
1.98
2.02
2.07
2.11
2.16
2.21
2.26
2.32
2.37
2.43
2.49
2.55
2.61
2.67
2.74
2.80
2.88
2.94
3.02
1.72
1.76
1.80
1.85
1.90
1.95
1.99
2.05
2.10
2.15
2.21
2.27
2.33
2.39
2.45
2.52
2.59
2.66
2.73
2.81
2.89
2.97
3.05
3.14
3.23
3.32
1.77
1.81
1.87
1.92
1.97
2.03
2.09
2.15
2.21
2.27
2.34
2.41
2.48
2.56
2.64
2.72
2.80
2.89
2.98
3.07
3.17
3.27
3.38
3.49
3.61
3.73
1.83
1.89
1.95
2.01
2.07
2.14
2.21
2.28
2.35
2.43
2.51
2.60
2.68
2.77
2.87
2.97
3.07
3.18
3.29
3.41
3.53
3.66
3.80
3.94
4.09
4.25
1.92
1.99
2.06
2.13
2.21
2.28
2.36
2.45
2.54
2.63
2.73
2.83
2.93
3.04
3.16
3.28
3.41
3.55
3.69
3.84
4.00
4.16
4.34
4.52
4.72
4.92
2.04
2.12
2.20
2.28
2.37
2.46
2.56
2.66
2.77
2.88
3.00
3.12
3.25
3.39
3.53
3.68
3.84
4.01
4.19
4.38
4.59
4.80
5.03
5.27
5.53
5.80
*
Based on Zhu and Qian, 2000
Table 7.12 Variation of R [Eq. (7.73)]
R for ␾⬘ (deg)
␪ (deg)
␦⬘/␾⬘
30
35
40
45
0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.65
1.95
2.2
1.2
1.4
1.6
1.9
2.15
1.28
1.6
1.95
2.4
2.85
1.25
1.6
1.9
2.35
2.8
1.35
1.8
2.4
3.15
3.95
1.32
1.8
2.35
3.05
3.8
1.45
2.2
3.2
4.45
6.1
1.4
2.1
3.0
4.3
5.7
5
370 Chapter 7: Lateral Earth Pressure
Table 7.12 (Continued)
R for ␾⬘ (deg)
␪ (deg)
␦⬘/␾⬘
30
35
40
45
10
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
0.2
0.4
0.6
0.8
1.0
1.15
1.35
1.6
1.8
2.05
1.15
1.35
1.55
1.8
2.0
1.15
1.35
1.5
1.8
1.9
1.2
1.5
1.85
2.25
2.65
1.2
1.5
1.8
2.2
2.6
1.2
1.45
1.8
2.1
2.4
1.3
1.7
2.25
2.9
3.6
1.3
1.65
2.2
2.8
3.4
1.3
1.65
2.1
2.6
3.2
1.4
2.0
2.9
4.0
5.3
1.35
1.95
2.7
3.8
4.95
1.35
1.9
2.6
3.55
4.8
15
20
7.14
Passive Pressure under Earthquake Conditions
The relationship for passive earth pressure on a retaining wall with a granular backfill and
under earthquake conditions was evaluated by Subba Rao and Choudhury (2005) by the
method of limit equilibrium using the pseudo-static approach. Figure 7.29 shows the
nature of failure surface in soil considered in this analysis. The passive pressure, Ppe,, can
be expressed as
Ppe 5 312gH 2Kpg(e) 4
1
cos dr
(7.74)
where Kpg(e) 5 passive earth-pressure coefficient in the normal direction to the wall.
A
D
45 /2
45 /2
H 2H
3
Ppe
C
Granular soil
c 0
B
Figure 7.29 Nature of failure surface in soil considered in the analysis
to determine Ppe
Problems
10
15
k 0
40
12
371
k kh
40
8
k 0
k kh
9
40
4
30
(e)
6
Kp
Kp
(e)
40
6
30
30
30
3
20
2
20
10
20
10
1
10
20
10
0.5
0
0
0
0.1
0.2
0.3
0.4
0.5
0
kh
(a)
Figure 7.30 Variation of Kpg(e) : (a)
0.1
0.2
0.3
0.4
0.5
kh
(b)
dr
dr
5 1, (b)
5 0.5
fr
fr
Kpg(e) is a function of kh and kv that are, respectively, coefficient of horizontal and
vertical acceleration due to earthquake. The variations of Kpg(e) for dr>fr 5 0.5 and 1 are
shown in Figures 7.30a and b. The passive pressure Ppe will be inclined at an angle d9
to the back face of the wall and will act at a distance of H>3 above the bottom of
the wall.
Problems
Refer to Figure 7.3a. Given: H 5 3.5 m, q 5 20 kN/m2, g 5 18.2 kN/m3 c9 5 0, and
f9 5 358. Determine the at-rest lateral earth force per meter length of the wall.
Also, find the location of the resultant. Use Eq. (7.4) and OCR = 1.5.
7.2 Use Eq. (7.3), Figure P7.2, and the following values to determine the at-rest lateral
earth force per unit length of the wall. Also find the location of the resultant.
H 5 5 m, H1 5 2 m, H2 5 3 m, g 5 15.5 kN>m3, gsat 5 18.5 kN>m3, f9 5 34°,
c9 5 0, q 5 20 kN>m2, and OCR 5 1.
7.3 Refer to Figure 7.6a. Given the height of the retaining wall, H is 6.4 m; the backfill
is a saturated clay with f 5 08, c 5 30.2 kN/m2, gsat 5 17.76 kN/m3,
a. Determine the Rankine active pressure distribution diagram behind the wall.
b. Determine the depth of the tensile crack, zc.
c. Estimate the Rankine active force per foot length of the wall before and after the
occurrence of the tensile crack.
7.1
372 Chapter 7: Lateral Earth Pressure
q
H1
z
1
c1
1
Groundwater
table
H
H2
2
c2
2
Figure P7.2
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
A vertical retaining wall (Figure 7.6a) is 6.3 m high with a horizontal backfill. For
the backfill, assume that g 5 17.9 kN>m3, f9 5 268, and c9 5 15 kN>m2. Determine
the Rankine active force per unit length of the wall after the occurrence of the tensile
crack.
Refer to Problem 7.2. For the retaining wall, determine the Rankine active
force per unit length of the wall and the location of the line of action of the
resultant.
Refer to Figure 7.10. For the retaining wall, H 5 6 m, fr 5 34°, a 5 10°,
g 5 17 kN>m3, and c9 5 0.
a. Determine the intensity of the Rankine active force at z 5 2 m, 4 m, and 6 m.
b. Determine the Rankine active force per meter length of the wall and also the
location and direction of the resultant.
Refer to Figure 7.10. Given: H 5 6.7 m, g 5 18.08 kN/m3, f9 5 258, c9 5 12 kN/m2,
and a 5 108. Calculate the Rankine active force per unit length of the wall after the
occurrence of the tensile crack.
Refer to Figure 7.12a. Given: H 5 3.66 m, g 5 16.5 kN/m3, f9 5 308, c9 5 0, and
b 5 858. Determine the Coulomb’s active force per foot length of the wall and the
location and direction of the resultant for the following cases:
a. a 5 108 and d9 5 208
b. a 5 208 and d9 5 158
Refer to Figure 7.13a. Given H ⫽ 3.5 m, ␣ ⫽ 0, ␤ ⫽ 85°, ␥ ⫽ 18 kN/m3, c⬘ ⫽ 0,
f9 5 348, ␦⬘/␾⬘ ⫽ 0.5, and q ⫽ 30 kN/m2. Determine the Coulomb’s active force
per unit length of the wall.
Refer to Figure 7.14b. Given H ⫽ 3.3 m, a⬘ ⫽ 1 m, b⬘ ⫽ 1.5 m, and q ⫽ 25 kN/m2.
Determine the lateral force per unit length of the unyielding wall caused by the surcharge loading only.
Refer to Figure 7.15. Here, H 5 6 m, g 5 17 kN>m3, f9 5 358, d9 5 17.58, c9 5 0,
a 5 108, and b 5 908. Determine the Coulomb’s active force for earthquake conditions (Pae) per meter length of the wall and the location and direction of the resultant. Given kh 5 0.2 and kv 5 0.
References
8m
373
= 17.5 kN/m3
= 35
c = 0
Pa
Figure P7.12
7.12 A retaining wall is shown in Figure P7.12. If the wall rotates about its top, determine
the magnitude of the active force per unit length of the wall for na 5 0.3, 0.4, and
0.5. Assume dr>fr 5 0.5.
7.13 A vertical frictionless retaining wall is 6-m high with a horizontal granular backfill.
Given: g 5 16 kN>m3 and f9 5 308. For the translation mode of the wall, calculate
the active pressure at depths z 5 1.5 m, 3 m, 4.5 m, and 6 m.
7.14 Refer to Problem 7.3.
a. Draw the Rankine passive pressure distribution diagram behind the wall.
b. Estimate the Rankine passive force per foot length of the wall and also the location of the resultant.
7.15 In Figure 7.28, which shows a vertical retaining wall with a horizontal backfill,
let H 5 4 m, u 5 108, g 5 16.5 kN>m3, f9 5 358, and d9 5 108. Based on
Zhu and Qian’s work, what would be the passive force per meter length of
the wall?
7.16 Consider a 4-m high retaining wall with a vertical back and horizontal granular backfill, as shown in Figure 7.29. Given: g 5 18 kN>m3, f9 5 408, c9 5 0, d9 5 208,
kv 5 0 and kh 5 0.2. Determine the passive force Ppe per unit length of the wall taking the earthquake effect into consideration.
References
CHU, S. C. (1991). “Rankine Analysis of Active and Passive Pressures on Dry Sand,” Soils and Foundations, Vol. 31, No. 4, pp. 115–120.
COULOMB, C. A. (1776). Essai sur une Application des Règles de Maximis et Minimum à quelques
Problemes de Statique Relatifs à l’Architecture, Mem. Acad. Roy. des Sciences, Paris,
Vol. 3, p. 38.
DUBROVA, G. A. (1963). “Interaction of Soil and Structures,” Izd. Rechnoy Transport, Moscow.
HARR, M. E. (1966). Fundamentals of Theoretical Soil Mechanics, McGraw-Hill, New York.
JAKY, J. (1944). “The Coefficient of Earth Pressure at Rest,” Journal for the Society of Hungarian
Architects and Engineers, October, pp. 355–358.
JARQUIO, R. (1981). “Total Lateral Surcharge Pressure Due to Strip Load,” Journal of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 107, No. GT10,
pp. 1424–1428.
KIM, J. S., and Preber, T. (1969). “Earth Pressure against Braced Excavations,” Journal of the Soil
Mechanics and Foundations Division, ASCE, Vol. 96, No. 6, pp. 1581–1584.
MAYNE, P. W., and Kulhawy, F. H. (1982). “Ko– OCR Relationships in Soil,” Journal of the Geotechnical Engineering Division, ASCE, Vol. 108, No. GT6, pp. 851– 872.
374 Chapter 7: Lateral Earth Pressure
MAZINDRANI, Z. H., and Ganjali, M. H. (1997). “Lateral Earth Pressure Problem of Cohesive Backfill with Inclined Surface,” Journal of Geotechnical and Geoenvironmental Engineering,
ASCE, Vol. 123, No. 2, pp. 110 –112.
SEED, H. B., and WHITMAN, R. V. (1970). “Design of Earth Retaining Structures for Dynamic
Loads,” Proceedings, Specialty Conference on Lateral Stresses in the Ground and Design of
Earth Retaining Structures, American Society of Civil Engineers, pp. 103–147.
SHIELDS, D. H., and Tolunay, A. Z. (1973). “Passive Pressure Coefficients by Method of Slices,” Journal
of the Soil Mechanics and Foundations Division, ASCE, Vol. 99, No. SM12, pp. 1043–1053.
SUBBA RAO, K. S., and CHOUDHURY, D. (2005). “Seismic Passive Earth Pressures in Soil,” Journal of
Geotechnical and Geoenvironmental Engineering, American Society of Civil Engineers,
Vo. 131, No. 1, pp. 131–135.
TERZAGHI, K. (1943). Theoretical Soil Mechanics, Wiley, New York.
ZHU, D. Y., and Qian, Q. (2000). “Determination of Passive Earth Pressure Coefficient by the Method
of Triangular Slices,” Canadian Geotechnical Journal, Vol. 37, No. 2, pp. 485–491.
8
8.1
Retaining Walls
Introduction
In Chapter 7, you were introduced to various theories of lateral earth pressure. Those theories will be used in this chapter to design various types of retaining walls. In general,
retaining walls can be divided into two major categories: (a) conventional retaining walls
and (b) mechanically stabilized earth walls.
Conventional retaining walls can generally be classified into four varieties:
1.
2.
3.
4.
Gravity retaining walls
Semigravity retaining walls
Cantilever retaining walls
Counterfort retaining walls
Gravity retaining walls (Figure 8.1a) are constructed with plain concrete or stone
masonry. They depend for stability on their own weight and any soil resting on the
masonry. This type of construction is not economical for high walls.
In many cases, a small amount of steel may be used for the construction of gravity
walls, thereby minimizing the size of wall sections. Such walls are generally referred to as
semigravity walls (Figure 8.1b).
Cantilever retaining walls (Figure 8.1c) are made of reinforced concrete that consists of a thin stem and a base slab. This type of wall is economical to a height of about
8 m. Figure 8.2 shows a cantilever retaining wall under construction.
Counterfort retaining walls (Figure 8.1d) are similar to cantilever walls. At regular
intervals, however, they have thin vertical concrete slabs known as counterforts that tie the
wall and the base slab together. The purpose of the counterforts is to reduce the shear and
the bending moments.
To design retaining walls properly, an engineer must know the basic parameters—
the unit weight, angle of friction, and cohesion—of the soil retained behind the wall
and the soil below the base slab. Knowing the properties of the soil behind the wall
enables the engineer to determine the lateral pressure distribution that has to be
designed for.
375
376 Chapter 8: Retaining Walls
Reinforcement
Reinforcement
Plain
concrete
or stone
masonry
(a) Gravity wall
(b) Semigravity wall
(c) Cantilever wall
Counterfort
(d) Counterfort wall
Figure 8.1 Types of retaining wall
There are two phases in the design of a conventional retaining wall. First, with the lateral earth pressure known, the structure as a whole is checked for stability. The structure is
examined for possible overturning, sliding, and bearing capacity failures. Second, each
component of the structure is checked for strength, and the steel reinforcement of each
component is determined.
This chapter presents the procedures for determining the stability of the retaining
wall. Checks for strength can be found in any textbook on reinforced concrete.
Some retaining walls have their backfills stabilized mechanically by including
reinforcing elements such as metal strips, bars, welded wire mats, geotextiles, and
8.2 Proportioning Retaining Walls
377
Figure 8.2 A cantilever retaining wall under construction (Courtesy of Dharma Shakya,
Geotechnical Solutions, Inc., Irvine, California)
geogrids. These walls are relatively flexible and can sustain large horizontal and vertical displacements without much damage.
Gravity and Cantilever Walls
8.2
Proportioning Retaining Walls
In designing retaining walls, an engineer must assume some of their dimensions. Called
proportioning, such assumptions allow the engineer to check trial sections of the walls for
stability. If the stability checks yield undesirable results, the sections can be changed and
rechecked. Figure 8.3 shows the general proportions of various retaining-wall components
that can be used for initial checks.
Note that the top of the stem of any retaining wall should not be less than about
0.3 m for proper placement of concrete. The depth, D, to the bottom of the base slab
should be a minimum of 0.6 m. However, the bottom of the base slab should be positioned below the seasonal frost line.
For counterfort retaining walls, the general proportion of the stem and the base slab
is the same as for cantilever walls. However, the counterfort slabs may be about 0.3 m thick
and spaced at center-to-center distances of 0.3H to 0.7H.
378
Chapter 8: Retaining Walls
0.3 m
min
0.3 m
min
min
0.02
min
0.02
I
I
H
H
Stem
0.1 H
D
0.12 to
Heel
0.17 H
Toe
0.12
to
0.17 H
D
0.1 H
0.1 H
0.5 to 0.7 H
0.5 to 0.7 H
(a)
(b)
Figure 8.3 Approximate dimensions for various components of retaining wall for initial stability
checks: (a) gravity wall; (b) cantilever wall
8.3
Application of Lateral Earth Pressure Theories
to Design
The fundamental theories for calculating lateral earth pressure were presented in Chapter 7.
To use these theories in design, an engineer must make several simple assumptions. In the
case of cantilever walls, the use of the Rankine earth pressure theory for stability checks
involves drawing a vertical line AB through point A, located at the edge of the heel of the
base slab in Figure 8.4a. The Rankine active condition is assumed to exist along the vertical plane AB. Rankine active earth pressure equations may then be used to calculate the
lateral pressure on the face AB of the wall. In the analysis of the wall’s stability, the force
Pa(Rankine) , the weight of soil above the heel, and the weight Wc of the concrete all should
be taken into consideration. The assumption for the development of Rankine active pressure along the soil face AB is theoretically correct if the shear zone bounded by the line
AC is not obstructed by the stem of the wall. The angle, h, that the line AC makes with the
vertical is
h 5 45 1
fr
1
sin a
a
2
2 sin21 ¢
≤
2
2
2
sin fr
(8.1)
8.3 Application of Lateral Earth Pressure Theories to Design
379
A similar type of analysis may be used for gravity walls, as shown in Figure 8.4b.
However, Coulomb’s active earth pressure theory also may be used, as shown in
Figure 8.4c. If it is used, the only forces to be considered are Pa(Coulomb) and the weight
of the wall, Wc .
␣
B
C
␥1
␾1
c1 0
H
Pa(Rankine)
␩
Ws
␥2
␾2
c2
H/3
Wc
A
(a)
B
␥1
␾1
c1 0
Pa(Rankine)
Ws
␥2
␾2
c2
Wc
A
(b)
Figure 8.4 Assumption for the determination of lateral earth
pressure: (a) cantilever wall; (b) and (c) gravity wall
380 Chapter 8: Retaining Walls
␥1
␾1
c1
Pa(Coulomb)
␦
Wc
␥2
␾2
c2
A
(c)
Figure 8.4 (continued)
If Coulomb’s theory is used, it will be necessary to know the range of the wall friction angle d9 with various types of backfill material. Following are some ranges of wall
friction angle for masonry or mass concrete walls:
Backfill material
Gravel
Coarse sand
Fine sand
Stiff clay
Silty clay
Range of d9 (deg)
27–30
20–28
15–25
15–20
12–16
In the case of ordinary retaining walls, water table problems and hence hydrostatic
pressure are not encountered. Facilities for drainage from the soils that are retained are
always provided.
8.4
Stability of Retaining Walls
A retaining wall may fail in any of the following ways:
•
•
•
•
•
It may overturn about its toe. (See Figure 8.5a.)
It may slide along its base. (See Figure 8.5b.)
It may fail due to the loss of bearing capacity of the soil supporting the base. (See
Figure 8.5c.)
It may undergo deep-seated shear failure. (See Figure 8.5d.)
It may go through excessive settlement.
The checks for stability against overturning, sliding, and bearing capacity failure will
be described in Sections 8.5, 8.6, and 8.7. The principles used to estimate settlement
8.4 Stability of Retaining Walls
(a)
381
(b)
(c)
Figure 8.5 Failure of retaining wall:
(a) by overturning; (b) by sliding;
(c) by bearing capacity failure;
(d) by deep-seated shear failure
(d)
Angle ␣ with
horizontal
f
a
O
For
␣ 10
d
c
e
b
Weak soil
Figure 8.6 Deep-seated shear failure
were covered in Chapter 5 and will not be discussed further. When a weak soil layer is
located at a shallow depth—that is, within a depth of 1.5 times the width of the base
slab of the retaining wall—the possibility of excessive settlement should be considered. In some cases, the use of lightweight backfill material behind the retaining wall
may solve the problem.
382 Chapter 8: Retaining Walls
Deep shear failure can occur along a cylindrical surface, such as abc shown in
Figure 8.6, as a result of the existence of a weak layer of soil underneath the wall at a depth
of about 1.5 times the width of the base slab of the retaining wall. In such cases, the critical cylindrical failure surface abc has to be determined by trial and error, using various
centers such as O. The failure surface along which the minimum factor of safety is
obtained is the critical surface of sliding. For the backfill slope with a less than about 10°,
the critical failure circle apparently passes through the edge of the heel slab (such as def in
the figure). In this situation, the minimum factor of safety also has to be determined by trial
and error by changing the center of the trial circle.
8.5
Check for Overturning
Figure 8.7 shows the forces acting on a cantilever and a gravity retaining wall, based on
the assumption that the Rankine active pressure is acting along a vertical plane AB drawn
through the heel of the structure. Pp is the Rankine passive pressure; recall that its magnitude is [from Eq. (7.63)].
Pp 5 12Kpg2D2 1 2c2r "KpD
where
g2 5 unit weight of soil in front of the heel and under the base slab
Kp 5 Rankine passive earth pressure coefficient 5 tan2 (45 1 f2r >2)
c2r , f2r 5 cohesion and effective soil friction angle, respectively
The factor of safety against overturning about the toe—that is, about point C in
Figure 8.7—may be expressed as
FS(overturning) 5
SMR
SMo
(8.2)
where
SMo 5 sum of the moments of forces tending to overturn about point C
SMR 5 sum of the moments of forces tending to resist overturning about point C
The overturning moment is
SMo 5 Ph ¢
Hr
≤
3
(8.3)
where Ph 5 Pa cos a .
To calculate the resisting moment, SMR (neglecting Pp ), a table such as Table 8.1 can
be prepared. The weight of the soil above the heel and the weight of the concrete (or
masonry) are both forces that contribute to the resisting moment. Note that the force Pv also
contributes to the resisting moment. Pv is the vertical component of the active force Pa , or
Pv 5 Pa sin a
The moment of the force Pv about C is
Mv 5 PvB 5 Pa sin aB
(8.4)
8.5 Check for Overturning
2
␣
383
A
␥1
␾1
c1 0
1
H
P␷
Pa
3
4
D
Ph
Pp
5
B
qheel
C
␥2
␾2
c2
qtoe
B
2
␣
A
␥1
␾1
c1 0
1
P␷
Pa
Ph
3
4
5
D
Pp
6
B
qheel
C
␥2
␾2
c2
qtoe
B
Figure 8.7 Check for overturning,
assuming that the Rankine pressure
is valid
where B 5 width of the base slab.
Once S MR is known, the factor of safety can be calculated as
FS(overturning) 5
M1 1 M2 1 M3 1 M4 1 M5 1 M6 1 Mv
Pa cos a(Hr>3)
(8.5)
The usual minimum desirable value of the factor of safety with respect to overturning is 2 to 3.
384 Chapter 8: Retaining Walls
Table 8.1 Procedure for Calculating SMR
Section
(1)
Area
(2)
1
2
3
4
5
6
A1
A2
A3
A4
A5
A6
Weight/unit
length of wall
(3)
Moment arm
measured from C
(4)
W1 5 g1 3 A 1
W2 5 g1 3 A 2
W3 5 gc 3 A 3
W4 5 gc 3 A 4
W5 5 gc 3 A 5
W6 5 gc 3 A 6
Pv
SV
X1
X2
X3
X4
X5
X6
B
Moment
about C
(5)
M1
M2
M3
M4
M5
M6
Mv
S MR
(Note: gl 5 unit weight of backfill
gc 5 unit weight of concrete)
Some designers prefer to determine the factor of safety against overturning with the
formula
M1 1 M2 1 M3 1 M4 1 M5 1 M6
FS(overturning) 5
(8.6)
Pa cos a(Hr>3) 2 Mv
8.6
Check for Sliding along the Base
The factor of safety against sliding may be expressed by the equation
FS(sliding) 5
S FRr
S Fd
(8.7)
where
S FRr 5 sum of the horizontal resisting forces
S Fd 5 sum of the horizontal driving forces
Figure 8.8 indicates that the shear strength of the soil immediately below the base
slab may be represented as
s 5 sr tan dr 1 car
where
dr 5 angle of friction between the soil and the base slab
car 5 adhesion between the soil and the base slab
Thus, the maximum resisting force that can be derived from the soil per unit length of the
wall along the bottom of the base slab is
Rr 5 s(area of cross section) 5 s(B 3 1) 5 Bsr tan dr 1 Bcar
However,
Bsr 5 sum of the vertical force 5 S V(see Table 8.1)
so
Rr 5 (S V) tan dr 1 Bcar
385
8.6 Check for Sliding along the Base
␥1
␾1
c1
V
Ph
D
Pp
␥2
␾2
c2
B
Figure 8.8 Check for sliding along the base
Figure 8.8 shows that the passive force Pp is also a horizontal resisting force. Hence,
S FRr 5 (S V) tan dr 1 Bcar 1 Pp
(8.8)
The only horizontal force that will tend to cause the wall to slide (a driving force) is
the horizontal component of the active force Pa , so
S Fd 5 Pa cos a
(8.9)
Combining Eqs. (8.7), (8.8), and (8.9) yields
FS(sliding) 5
(S V) tan dr 1 Bcra 1 Pp
Pa cos a
(8.10)
A minimum factor of safety of 1.5 against sliding is generally required.
In many cases, the passive force Pp is ignored in calculating the factor of safety with
respect to sliding. In general, we can write dr 5 k1f2r and car 5 k2c2r . In most cases, k1 and
k2 are in the range from 12 to 23 . Thus,
FS(sliding) 5
(S V) tan (k1f2r ) 1 Bk2c2r 1 Pp
Pa cos a
(8.11)
If the desired value of FS(sliding) is not achieved, several alternatives may be investigated
(see Figure 8.9):
• Increase the width of the base slab (i.e., the heel of the footing).
• Use a key to the base slab. If a key is included, the passive force per unit length of
the wall becomes
Pp 5
where Kp 5 tan2 ¢ 45 1
1
g D2K 1 2c2r D1"Kp
2 2 1 p
f2r
≤.
2
386 Chapter 8: Retaining Walls
Use of a dead
man anchor
␥1
␾1
c1
D
D1
Base slab
increase
Use of a
base key
Pp
␥2
␾2
c2
Figure 8.9 Alternatives for increasing the factor of safety with respect to sliding
• Use a deadman anchor at the stem of the retaining wall.
• Another possible way to increase the value of FS(sliding) is to consider reducing the
value of Pa [see Eq. (8.11)]. One possible way to do so is to use the method developed by Elman and Terry (1988). The discussion here is limited to the case in which
the retaining wall has a horizontal granular backfill (Figure 8.10). In Figure 8.10,
the active force, Pa, is horizontal (␣ ⫽ 0) so that
Pacos ␣ ⫽ Ph ⫽ Pa
and
Pa sin ␣ ⫽ Pv ⫽ 0
However,
Pa ⫽ Pa(1) ⫹ Pa(2)
(8.12)
The magnitude of Pa(2) can be reduced if the heel of the retaining wall is sloped
as shown in Figure 8.10. For this case,
Pa ⫽ Pa(1) ⫹ APa(2)
(8.13)
The magnitude of A, as shown in Table 8.2, is valid for ␣⬘ ⫽ 45°. However
note that in Figure 8.10a
Pa(1) 5
1
g K (Hr 2 Dr ) 2
2 1 a
and
Pa 5
1
g K Hr2
2 1 a
Hence,
Pa(2) 5
1
g K 3Hr2 2 (H 9 2 D 9 ) 24
2 1 a
387
8.7 Check for Bearing Capacity Failure
␥1
␾1
c1= 0
␥1
␾1
c1= 0
H
Pa
= [Eq.
(8.12)]
Pa(1)
D
D
Pa(1)
Pa
= [Eq.
(8.13)]
APa(2)
Pa(2)
α
(a)
(b)
Figure 8.10 Retaining wall with sloped heel
Table 8.2 Variation of A with ␾1⬘ (for ␣⬘ ⫽ 45°)
Soil friction
angle, ␾1ⴕ (deg)
A
20
25
30
35
40
0.28
0.14
0.06
0.03
0.018
So, for the active pressure diagram shown in Figure 8.10b,
Pa 5
1
A
g1Ka (H 9 2 D 9 ) 2 1 g1Ka 3H 92 2 (H 9 2 D 9 ) 24
2
2
(8.14)
Sloping the heel of a retaining wall can thus be extremely helpful in some cases.
8.7
Check for Bearing Capacity Failure
The vertical pressure transmitted to the soil by the base slab of the retaining wall should
be checked against the ultimate bearing capacity of the soil. The nature of variation of the
vertical pressure transmitted by the base slab into the soil is shown in Figure 8.11. Note
that qtoe and qheel are the maximum and the minimum pressures occurring at the ends of the
toe and heel sections, respectively. The magnitudes of qtoe and qheel can be determined in
the following manner:
The sum of the vertical forces acting on the base slab is S V (see column 3 of
Table 8.1), and the horizontal force Ph is Pa cos a. Let
R 5 S V 1 Ph
(8.15)
be the resultant force. The net moment of these forces about point C in Figure 8.11 is
Mnet 5 SMR 2 SMo
(8.16)
388 Chapter 8: Retaining Walls
␥1
␾1
c1 0
V
V
R
Ph Pa cos ␣
X
␥2
␾2
c2
D
Ph Ph
C
E
qmax qtoe
qmin qheel
e
B/2
y
B/2
Figure 8.11 Check for bearing
capacity failure
Note that the values of SMR and SMo were previously determined. [See Column 5 of
Table 8.1 and Eq. (8.3)]. Let the line of action of the resultant R intersect the base slab at E.
Then the distance
CE 5 X 5
Mnet
SV
(8.17)
Hence, the eccentricity of the resultant R may be expressed as
e5
B
2 CE
2
(8.18)
The pressure distribution under the base slab may be determined by using simple
principles from the mechanics of materials. First, we have
q5
Mnety
SV
6
A
I
where
Mnet 5 moment 5 (SV)e
I 5 moment of inertia per unit length of the base section
5 121 (1) (B3 )
(8.19)
8.7 Check for Bearing Capacity Failure
389
For maximum and minimum pressures, the value of y in Eq. (8.19) equals B>2. Substituting into Eq. (8.19) gives
qmax 5 qtoe 5
SV
1
(B) (1)
e(SV)
B
2
1
¢ ≤ (B3 )
12
5
SV
6e
¢1 1 ≤
B
B
(8.20)
Similarly,
qmin 5 qheel 5
SV
6e
¢1 2 ≤
B
B
(8.21)
Note that SV includes the weight of the soil, as shown in Table 8.1, and that when the
value of the eccentricity e becomes greater than B>6, qmin [Eq. (8.21)] becomes negative.
Thus, there will be some tensile stress at the end of the heel section. This stress is not desirable, because the tensile strength of soil is very small. If the analysis of a design
shows that e . B>6, the design should be reproportioned and calculations redone.
The relationships pertaining to the ultimate bearing capacity of a shallow foundation
were discussed in Chapter 3. Recall that [Eq. (3.40)].
qu 5 c2r NcFcdFci 1 qNqFqdFqi 1 12g2BrNgFgdFgi
(8.22 )
where
q 5 g2D
Br 5 B 2 2e
1 2 Fqd
Fcd 5 Fqd 2
Nc tan f2r
Fqd 5 1 1 2 tan f2r (1 2 sin f2r ) 2
D
Br
Fgd 5 1
Fci 5 Fqi 5 ¢ 1 2
Fgi 5 ¢1 2
c° 2
≤
90°
c° 2
≤
f2r °
c° 5 tan21 ¢
Pa cos a
≤
SV
Note that the shape factors Fcs , Fqs , and Fgs given in Chapter 3 are all equal to unity, because they can be treated as a continuous foundation. For this reason, the shape factors are
not shown in Eq. (8.22).
Once the ultimate bearing capacity of the soil has been calculated by using
Eq. (8.22), the factor of safety against bearing capacity failure can be determined:
FS(bearing capacity) 5
qu
qmax
(8.23)
Generally, a factor of safety of 3 is required. In Chapter 3, we noted that the ultimate bearing
capacity of shallow foundations occurs at a settlement of about 10% of the foundation width.
390 Chapter 8: Retaining Walls
In the case of retaining walls, the width B is large. Hence, the ultimate load qu will occur at
a fairly large foundation settlement. A factor of safety of 3 against bearing capacity failure
may not ensure that settlement of the structure will be within the tolerable limit in all cases.
Thus, this situation needs further investigation.
An alternate relationship to Eq. (8.22) will be Eq. (3.67), or
qu 5 crNc(ei)Fcd 1 qNq(ei)Fqd 1 12g2BNg(ei)Fgd
Since Fgd 5 1,
qu 5 crNc(ei)Fcd 1 qNq(ei)Fqd 1 12g2BNg(ei)
(8.24)
The bearing capacity factors, Nc(ei), Nq(ei), and Ng(ei) were given in Figures 3.26
through 3.28.
Example 8.1
The cross section of a cantilever retaining wall is shown in Figure 8.12. Calculate the
factors of safety with respect to overturning, sliding, and bearing capacity.
10
0.5 m
H1 = 0.458 m
5
1 = 18 kN/m3
1= 30
c1= 0
H2 = 6 m
1
Pa
Pv
10
4
Ph
2
1.5 m = D
0.7 m
3
H3 = 0.7 m
C
0.7 m
0.7 m
2.6 m
Figure 8.12 Calculation of stability of a retaining wall
␥2 = 19 kN/m3
␾2 = 20
c2 = 40 kN/m2
8.7 Check for Bearing Capacity Failure
391
Solution
From the figure,
H⬘ ⫽ H1 ⫹ H2 ⫹ H3 ⫽ 2.6 tan 10° ⫹ 6 ⫹ 0.7
⫽ 0.458 ⫹ 6 ⫹ 0.7 ⫽ 7.158 m
The Rankine active force per unit length of wall ⫽ Pp 5 12g1Hr2Ka . For ␾1⬘ ⫽ 30° and
␣ ⫽ 10°, Ka is equal to 0.3532. (See Table 7.1.) Thus,
Pa 5 12 (18) (7.158) 2 (0.3532) 5 162.9 kN>m
Pv ⫽ Pa sin10° ⫽ 162.9 (sin10°) ⫽ 28.29 kN/m
and
Ph ⫽ Pa cos10° ⫽ 162.9 (cos10°) ⫽ 160.43 kN/m
Factor of Safety against Overturning
The following table can now be prepared for determining the resisting moment:
Section
no.a
Area
(m2)
Weight/unit
length
(kN/m)
Moment arm
from point C
(m)
70.74
14.15
1.15
0.833
66.02
280.80
10.71
Pv ⫽ 28.29
⌺V ⫽ 470.71
2.0
2.7
3.13
4.0
6 ⫻ 0.5 ⫽ 3
5 0.6
1
2
1
2 (0.2)6
3
4
5
4 ⫻ 0.7 ⫽ 2.8
6 ⫻ 2.6 ⫽ 15.6
1
2 (2.6) (0.458) 5 0.595
Moment
(kN-m/m)
81.35
11.79
132.04
758.16
33.52
113.16
1130.02 ⫽ ⌺MR
a
For section numbers, refer to Figure 8.12
␥concrete ⫽ 23.58 kN/m3
The overturning moment
Mo 5 Ph a
7.158
H9
b 5 160.43a
b 5 382.79 kN-m>m
3
3
and
FS (overturning) 5
SMR
1130.02
5
5 2.95 . 2, OK
Mo
382.79
Factor of Safety against Sliding
From Eq. (8.11),
FS (sliding) 5
(SV)tan(k1fr2 ) 1 Bk2cr2 1 Pp
Pacosa
392 Chapter 8: Retaining Walls
Let k1 5 k2 5 32 . Also,
Pp 5 12Kpg2D2 1 2cr2 !KpD
Kp 5 tan2 a45 1
fr2
b 5 tan2 (45 1 10) 5 2.04
2
and
D ⫽ 1.5 m
So
Pp 5 12 (2.04) (19) (1.5) 2 1 2(40) ( !2.04) (1.5)
⫽ 43.61 ⫹ 171.39 ⫽ 215 kN/m
Hence,
(470.71)tana
FS (sliding) 5
5
2 3 20
2
b 1 (4) a b (40) 1 215
3
3
160.43
111.56 1 106.67 1 215
5 2.7 > 1.5, OK
160.43
Note: For some designs, the depth D in a passive pressure calculation may be taken to
be equal to the thickness of the base slab.
Factor of Safety against Bearing Capacity Failure
Combining Eqs. (8.16), (8.17), and (8.18) yields
SMR 2 SMo
B
4
1130.02 2 382.79
2
5 2
2
SV
2
470.71
B
4
5 0.411 m , 5 5 0.666 m
6
6
e5
Again, from Eqs. (8.20) and (8.21)
q toe
heel 5
SV
6e
470.71
6 3 0.411
a1 6 b 5
a1 6
b 5 190.2 kN>m2 (toe)
B
B
4
4
5 45.13 kN>m2 (heel)
The ultimate bearing capacity of the soil can be determined from Eq. (8.22)
qu 5 c29NcFcdFci 1 qNqFqdFqi 1
1
g B9NgFgdFgi
2 2
8.7 Check for Bearing Capacity Failure
393
For ␾2⬘ ⫽ 20° (see Table 3.3), Nc ⫽ 14.83, Nq ⫽ 6.4, and N␥ ⫽ 5.39. Also,
q ⫽ ␥2D ⫽ (19) (1.5) ⫽ 28.5 kN/m2
B⬘ ⫽ B ⫺ 2e ⫽ 4 ⫺ 2(0.411) ⫽ 3.178 m
1 2 Fqd
1 2 1.148
Fcd 5 Fqd 2
5 1.148 2
5 1.175
Nctanfr2
(14.83) (tan 20)
2
Fqd 5 1 1 2 tanfr2 (1 2 sinfr)
2 a
D
1.5
b 5 1 1 0.315a
b 5 1.148
Br
3.178
F␥d ⫽ 1
Fci 5 Fqi 5 a1 2
c° 2
b
90°
and
c 5 tan21 a
Pacosa
160.43
b 5 tan21 a
b 5 18.82°
SV
470.71
So
Fci 5 Fqi 5 a1 2
18.82 2
b 5 0.626
90
and
Fgi 5 a1 2
c 2
18.82 2
b 5 a1 2
b <0
f2r
20
Hence,
qu 5 (40) (14.83) (1.175) (0.626) 1 (28.5) (6.4) (1.148) (0.626)
1 12 (19) (5.93) (3.178) (1) (0)
⫽ 436.33 ⫹ 131.08 ⫹ 0 ⫽ 567.41 kN/m2
and
FS (bearing capacity) 5
qu
567.41
5
5 2.98
q toe
190.2
Note: FS(bearing capacity) is less than 3. Some repropertioning will be needed.
■
Example 8.2
A gravity retaining wall is shown in Figure 8.13. Use dr 5 2>3f1r and Coulomb’s active
earth pressure theory. Determine
394 Chapter 8: Retaining Walls
␥1 18.5 kN/m3
␾1 32°
c1 0
P␷
5.7 m
5m
2
Pa
1
␦
15°
2.83 m
Ph
3
75°
2.167 m
1.5 m
0.27 m 0.6 m
0.8 m
1.53 m
4
C
0.3 m
0.8 m
3.5 m
␥2 18 kN/m3
␾2 24°
c2 30 kN/m2
Figure 8.13 Gravity retaining wall (not to scale)
a. The factor of safety against overturning
b. The factor of safety against sliding
c. The pressure on the soil at the toe and heel
Solution
The height
Hr 5 5 1 1.5 5 6.5 m
Coulomb’s active force is
Pa 5 12 g1Hr2Ka
With a 5 0°, b 5 75°, dr 5 2>3f1r , and f1r 5 32°, Ka 5 0.4023. (See Table 7.4.) So,
Pa 5 12 (18.5) (6.5) 2 (0.4023) 5 157.22 kN>m
Ph 5 Pa cos (15 1 23f1r ) 5 157.22 cos 36.33 5 126.65 kN>m
and
Pv 5 Pa sin (15 1 23f1r ) 5 157.22 sin 36.33 5 93.14 kN>m
8.7 Check for Bearing Capacity Failure
395
Part a: Factor of Safety against Overturning
From Figure 8.13, one can prepare the following table:
Area
no.
1
2
3
4
*
1
2 (5.7)
Moment arm
from C
(m)
Weight*
(kN/m)
Area
(m2 )
(1.53) 5 4.36
(0.6) (5.7) 5 3.42
1
2 (0.27) (5.7) 5 0.77
< (3.5) (0.8) 5 2.8
102.81
80.64
18.16
66.02
Pv 5 93.14
SV 5 360.77 kN>m
2.18
1.37
0.98
1.75
2.83
Moment
(kN-m/m)
224.13
110.48
17.80
115.54
263.59
SMR 5 731.54 kN-m>m
gconcrete 5 23.58 kN>m3
Note that the weight of the soil above the back face of the wall is not taken into account
in the preceding table. We have
Overturning moment 5 Mo 5 Ph ¢
Hr
≤ 5 126.65(2.167) 5 274.45 kN-m>m
3
Hence,
FS(overturning) 5
SMR
731.54
5
5 2.67 + 2, OK
SMo
274.45
Part b: Factor of Safety against Sliding
We have
FS(sliding) 5
2
2
(SV) tan ¢ f2r ≤ 1 c2r B 1 Pp
3
3
Ph
Pp 5 12Kpg2D2 1 2c2r "KpD
and
Kp 5 tan2 ¢45 1
24
≤ 5 2.37
2
Hence,
Pp 5 12 (2.37) (18) (1.5) 2 1 2(30) (1.54) (1.5) 5 186.59 kN>m
So
FS(sliding) 5
2
2
360.77 tan ¢ 3 24≤ 1 (30) (3.5) 1 186.59
3
3
126.65
396 Chapter 8: Retaining Walls
5
103.45 1 70 1 186.59
5 2.84
126.65
If Pp is ignored, the factor of safety is 1.37.
Part c: Pressure on Soil at Toe and Heel
From Eqs. (8.16), (8.17), and (8.18),
e5
qtoe 5
SMR 2 SMo
B
3.5
731.54 2 274.45
B
2
5
2
5 0.483 , 5 0.583
2
SV
2
360.77
6
(6) (0.483)
SV
6e
360.77
B1 1 R 5
B1 1
R 5 188.43 kN , m2
B
B
3.5
3.5
and
qheel 5
8.8
(6) (0.483)
V
6e
360.77
B1 2 R 5
B1 2
R 5 17.73 kN , m2
B
B
3.5
3.5
■
Construction Joints and Drainage from Backfill
Construction Joints
A retaining wall may be constructed with one or more of the following joints:
1. Construction joints (see Figure 8.14a) are vertical and horizontal joints that are
placed between two successive pours of concrete. To increase the shear at the joints,
keys may be used. If keys are not used, the surface of the first pour is cleaned and
roughened before the next pour of concrete.
2. Contraction joints (Figure 8.14b) are vertical joints (grooves) placed in the face
of a wall (from the top of the base slab to the top of the wall) that allow the concrete to shrink without noticeable harm. The grooves may be about
6 to 8 mm wide and 12 to 16 mm deep.
3. Expansion joints (Figure 8.14c) allow for the expansion of concrete caused by
temperature changes; vertical expansion joints from the base to the top of the wall
may also be used. These joints may be filled with flexible joint fillers. In most
cases, horizontal reinforcing steel bars running across the stem are continuous
through all joints. The steel is greased to allow the concrete to expand.
Drainage from the Backfill
As the result of rainfall or other wet conditions, the backfill material for a retaining wall
may become saturated, thereby increasing the pressure on the wall and perhaps creating
an unstable condition. For this reason, adequate drainage must be provided by means of
weep holes or perforated drainage pipes. (See Figure 8.15.)
When provided, weep holes should have a minimum diameter of about 0.1 m and
be adequately spaced. Note that there is always a possibility that backfill material may
8.8 Construction Joints and Drainage from Backfill
397
Roughened
surface
Keys
(a)
Back of wall
Back of wall
Contraction
joint
Face of wall
(b)
Expansion
joint
(c)
Face of
wall
Figure 8.14 (a) Construction joints; (b) contraction joint; (c) expansion joint
Weep hole
Filter material
Filter material
Perforated pipe
(a)
(b)
Figure 8.15 Drainage provisions for the backfill of a retaining wall: (a) by weep holes;
(b) by a perforated drainage pipe
be washed into weep holes or drainage pipes and ultimately clog them. Thus, a filter material needs to be placed behind the weep holes or around the drainage pipes, as the case may
be; geotextiles now serve that purpose.
Two main factors influence the choice of filter material: The grain-size distribution
of the materials should be such that (a) the soil to be protected is not washed into the filter and (b) excessive hydrostatic pressure head is not created in the soil with a lower
398 Chapter 8: Retaining Walls
hydraulic conductivity (in this case, the backfill material). The preceding conditions can
be satisfied if the following requirements are met (Terzaghi and Peck, 1967):
D15(F)
D85(B)
D15(F)
D15(B)
,5
3to satisfy condition(a)4
(8.25)
.4
3to satisfy condition(b) 4
(8.26)
In these relations, the subscripts F and B refer to the filter and the base material (i.e., the
backfill soil), respectively. Also, D15 and D85 refer to the diameters through which 15%
and 85% of the soil (filter or base, as the case may be) will pass. Example 8.3 gives the procedure for designing a filter.
Example 8.3
Figure 8.16 shows the grain-size distribution of a backfill material. Using the conditions outlined in Section 8.8, determine the range of the grain-size distribution for the filter material.
100
Range for
filter
material
Percent finer
80
D85(B)
Backfill
material
60
D50(B)
25 D50(B)
40
20
5 D85(B)
4 D15(B)
D15(B)
20 D15(B)
0
10
5
2
1
0.5
0.2 0.1
Grain size (mm)
0.05
0.02
0.01
Figure 8.16 Determination of grain-size distribution of filter material
Solution
From the grain-size distribution curve given in the figure, the following values can be
determined:
D15(B) 5 0.04 mm
D85(B) 5 0.25 mm
D50(B) 5 0.13 mm
8.9 Gravity Retaining-Wall Design for Earthquake Conditions
399
Conditions of Filter
1.
2.
3.
4.
D15(F)
D15(F)
D50(F)
D15(F)
should be less than 5D85(B) ; that is, 5 3 0.25 5 1.25 mm.
should be greater than 4D15(B) ; that is, 4 3 0.04 5 0.16 mm.
should be less than 25D50(B) ; that is, 25 3 0.13 5 3.25 mm.
should be less than 20D15(B) ; that is, 20 3 0.04 5 0.8 mm.
These limiting points are plotted in Figure 8.16. Through them, two curves can be
drawn that are similar in nature to the grain-size distribution curve of the backfill material. These curves define the range of the filter material to be used.
■
8.9
Gravity Retaining-Wall Design
for Earthquake Conditions
Even in mild earthquakes, most retaining walls undergo limited lateral displacement.
Richards and Elms (1979) proposed a procedure for designing gravity retaining walls for
earthquake conditions that allows limited lateral displacement. This procedure takes into
consideration the wall inertia effect. Figure 8.17 shows a retaining wall with various forces
acting on it, which are as follows (per unit length of the wall):
a. Ww ⫽ weight of the wall
b. Pae ⫽ active force with earthquake condition taken into consideration (Section 7.7)
The backfill of the wall and the soil on which the wall is resting are assumed cohesionless. Considering the equilibrium of the wall, it can be shown that
Ww 5 C12g1H 2 (1 2 kv )KaeD CIE
(8.27)
a
␥1
␾1
c1 = 0
Pae
␦
90 ⫺ ␤
kvWw
H
khWw
Ww
S = khWw Pae sin(␤
N = Ww
␦⬘)
kvWw Pae cos(␤
␦⬘)
z
␤
␥2
␾2
c2= 0
Figure 8.17 Stability
of a retaining wall under
earthquake forces
400 Chapter 8: Retaining Walls
where ␥1 ⫽ unit weight of the backfill;
CIE 5
and u 9 5 tan21 a
sin(b 2 d9 ) 2 cos(b 2 d9 )tanf29
(1 2 kv ) (tanf29 2 tan u9)
(8.28)
kk
b
1 2 kv
For a detailed derivation of Eq. (8.28), see Das (1983).
Based on Eqs. (8.27) and (8.28), the following procedure may be used to determine
the weight of the retaining wall, Ww, for tolerable displacement that may take place during an earthquake.
1. Determine the tolerable displacement of the wall, ⌬.
2. Obtain a design value of kk from
kk 5 A a a
0.2A 2v 0.25
b
AaD
(8.29)
In Eq. (8.29), A and Aa are effective acceleration coefficients and ⌬ is displacement
in inches. The magnitudes of Aa and Av are given by the Applied Technology
Council (1978) for various regions of the United States
3. Assume that kv ⫽ 0, and, with the value of kk obtained, calculate Kae from Eq. (7.43).
4. Use the value of Kae determined in Step 3 to obtain the weight of the wall (Ww)
5. Apply a factor of safety to the value of Ww obtained in Step 4.
Example 8.4
Refer to Figure 8.18. For kv ⫽ 0 and kk ⫽ 0.3, determine:
a. Weight of the wall for static condition
b. Weight of the wall for zero displacement during an earthquake
c. Weight of the wall for lateral displacement of 38 mm (1.5 in.) during an earthquake
␾1 = 36
␥1 = 16 kN/m3
␦ = 2/3 ␾1
5m
␾2 = 36
␥2 = 16 kN/m3
Figure 8.18
8.9 Gravity Retaining-Wall Design for Earthquake Conditions
401
For part c, assume that Aa ⫽ 0.2 and Av ⫽ 0.2. For parts a, b, and c, use a factor of
safety of 1.5.
Solution
Part a
For static conditions, ␪⬘ ⫽ 0 and Eq. (8.28) becomes
CIE 5
sin(b 2 dr ) 2 cos(b 2 dr )tanf2r
tanf2r
For ␤ ⫽ 90°, ␦⬘ ⫽ 24° and ␾⬘2 ⫽ 36°,
CIE 5
sin(90 2 24) 2 cos(90 2 24)tan 36
5 0.85
tan 36
For static conditions, Kae ⫽ Ka, so
1
Ww 5 gH 2KaCIE
2
For Ka ⬇ 0.2349 [Table 7.4],
1
Ww 5 (16) (5) 2 (0.2349) (0.85) 5 39.9 kN>m
2
With a factor of safety of 1.5,
Ww ⫽ (39.9)(1.5) ⫽ 59.9 kN/m
Part b
For zero displacement, kv ⫽ 0,
CIE 5
tan ur 5
CIE 5
sin(b 2 dr ) 2 cos(b 2 dr )tan f2r
tan f2r 2 tan ur
kh
0.3
5
5 0.3
1 2 kv
120
sin(90 2 24) 2 cos(90 2 24)tan 36
5 1.45
tan 36 2 0.3
For kh ⫽ 0.3, ␾1⬘ ⫽ 36° and ␦⬘ ⫽ 2␾1⬘/3, the value of Kae ⬇ 0.48 (Table 7.6).
Ww 5 12g1H 2 (1 2 kv )KaeCIE 5 12 (16) (5) 2 (1 2 0) (0.48) (1.45) 5 139.2 kN>m
With a factor of safety of 1.5, Ww ⫽ 208.8 kN/m
Part c
For a lateral displacement of 38 mm,
kh 5 A a a
0.2A 2v 0.25
(0.2) (0.2) 2 0.25
b
5 (0.2) c
d
5 0.081
AaD
(0.2) (38>25.4)
402 Chapter 8: Retaining Walls
tan ur 5
CIE 5
kh
0.081
5
5 0.081
1 2 kv
120
sin (90 2 24) 2 cos (90 2 24)tan 36
5 0.957
tan 36 2 0.081
1
g H 2KaeClE
2 1
h
Ww 5
⬇ 0.29 [Table 7.6]
1
Ww 5 (16) (5) 2 (0.29) (0.957) 5 55.5 kN>m
2
With a factor of safety of 1.5, Ww ⫽ 83.3 kN/m
8.10
■
Comments on Design of Retaining Walls
and a Case Study
In Section 8.3, it was suggested that the active earth pressure coefficient be used to estimate the lateral force on a retaining wall due to the backfill. It is important to recognize
the fact that the active state of the backfill can be established only if the wall yields sufficiently, which does not happen in all cases. The degree to which the wall yields depends
on its height and the section modulus. Furthermore, the lateral force of the backfill depends
on several factors identified by Casagrande (1973):
1.
2.
3.
4.
5.
6.
7.
Effect of temperature
Groundwater fluctuation
Readjustment of the soil particles due to creep and prolonged rainfall
Tidal changes
Heavy wave action
Traffic vibration
Earthquakes
Insufficient wall yielding combined with other unforeseen factors may generate a
larger lateral force on the retaining structure, compared with that obtained from the active
earth-pressure theory. This is particularly true in the case of gravity retaining walls, bridge
abutments, and other heavy structures that have a large section modulus.
Case Study for the Performance of a Cantilever Retaining Wall
Bentler and Labuz (2006) have reported the performance of a cantilever retaining wall
built along Interstate 494 in Bloomington, Minnesota. The retaining wall had 83 panels,
each having a length of 9.3 m. The panel height ranged from 4.0 m to 7.9 m. One of the
7.9 m high panels was instrumented with earth pressure cells, tiltmeters, strain gauges, and
8.10 Comments on Design of Retaining Walls and a Case Study
403
Granular
backfill (SP)
␥1 = 18.9 kN/m3
␾1= 35 to 39
(Av. 37)
2.4
7.9 m
Figure 8.19 Schematic
diagram of the retaining wall
(drawn to scale)
Poorly graded
sand, and sand and
gravel
inclinometer casings. Figure 8.19 shows a schematic diagram (cross section) of the wall
panel. Some details on the backfill and the foundation material are:
•
Granular Backfill
Effective size, D10 ⫽ 0.13 mm
Uniformity coefficient, Cu ⫽ 3.23
Coefficient of gradation, Cc ⫽ 1.4
Unified soil classification ⫺ SP
Compacted unit weight, ␥1 ⫽ 18.9 kN/m3
Triaxial friction angle, ␾1⬘ ⫺ 35° to 39° (average 37°)
•
Foundation Material
Poorly graded sand and sand with gravel (medium dense to dense)
The backfill and compaction of the granular material started on October 28, 2001 in
stages and reached a height of 7.6 m on November 21, 2001. The final 0.3 m of soil was
placed the following spring. During backfilling, the wall was continuously going through
translation (see Section 7.9). Table 8.3 is a summary of the backfill height and horizontal
translation of the wall.
Table 8.3 Horizontal Translation with Backfill Height
Day
Backfill height (m)
Horizontal translation (mm)
1
2
2
3
4
5
11
24
54
0.0
1.1
2.8
5.2
6.1
6.4
6.7
7.3
7.6
0
0
0
2
4
6
9
12
11
404 Chapter 8: Retaining Walls
Height of fill
above footing (m)
8
6.1 m
6
Observed pressure
Rankine active
pressure
4
(␾⬘1 ⬇ 37)
2
0
0
10
20
30
40
50
Lateral pressure (kN/m2)
Figure 8.20 Observed lateral pressure distribution after fill height
reached 6.1 m (Based on Bentler
and Labuz, 2006)
Figure 8.20 shows a typical plot of the variation of lateral earth pressure after compaction, ␴a⬘, when the backfill height was 6.1 m (October 31, 2001) along with the plot of
Rankine active earth pressure (␾1⬘ ⫽ 37°). Note that the measured lateral (horizontal) pressure is higher at most heights than that predicted by the Rankine active pressure theory,
which may be due to residual lateral stresses caused by compaction. The measured lateral
stress gradually reduced with time. This is demonstrated in Figure 8.21 which shows a plot
of the variation of ␴a⬘ with depth (November 27, 2001) when the height of the backfill was
7.6 m. The lateral pressure was lower at practically all depths compared to the Rankine
active earth pressure.
Another point of interest is the nature of variation of qmax and qmin (see Figure 8.11).
As shown in Figure 8.11, if the wall rotates about C, qmax will be at the toe and qmin will be
at the heel. However, for the case of the retaining wall under consideration (undergoing
Height of fill
above footing (m)
8
7.6 m
6
Rankine
active pressure
(␾1 ⬇ 37)
4
2
0
Observed
pressure
0
10
20
30
40
Lateral pressure (kN/m2)
50
Figure 8.21 Observed pressure
distribution on November 27,
2001 (Based on Bentler and
Labuz, 2006)
8.11 Soil Reinforcement
405
horizontal translation), qmax was at the heel of the wall with qmin at the toe. On November 27,
2001, when the height of the fill was 7.6 m, qmax at the heel was about 140 kN/m2, which
was approximately equal to (␥1)(height of fill) ⫽(18.9)(7.6) ⫽ 143.6 kN/m2. Also, at the
toe, qmin was about 40 kN/m2, which suggests that the moment from lateral force had little
effect on the vertical effective stress below the heel.
The lessons learned from this case study are the following:
a. Retaining walls may undergo lateral translation which will affect the variation of qmax
and qmin along the base slab.
b. Initial lateral stress caused by compaction gradually decreases with time and lateral
movement of the wall.
Mechanically Stabilized Retaining Walls
More recently, soil reinforcement has been used in the construction and design of foundations,
retaining walls, embankment slopes, and other structures. Depending on the type of construction, the reinforcements may be galvanized metal strips, geotextiles, geogrids, or geocomposites. Sections 8.11 and 8.12 provide a general overview of soil reinforcement and various
reinforcement materials.
Reinforcement materials such as metallic strips, geotextiles, and geogrids are now
being used to reinforce the backfill of retaining walls, which are generally referred to as
mechanically stabilized retaining walls. The general principles for designing these walls
are given in the following sections.
8.11
Soil Reinforcement
The use of reinforced earth is a recent development in the design and construction of foundations and earth-retaining structures. Reinforced earth is a construction material made from soil
that has been strengthened by tensile elements such as metal rods or strips, nonbiodegradable
fabrics (geotextiles), geogrids, and the like. The fundamental idea of reinforcing soil is not new;
in fact, it goes back several centuries. However, the present concept of systematic analysis and
design was developed by a French engineer, H. Vidal (1966). The French Road Research
Laboratory has done extensive research on the applicability and the beneficial effects of the use
of reinforced earth as a construction material. This research has been documented in detail by
Darbin (1970), Schlosser and Long (1974), and Schlosser and Vidal (1969). The tests that were
conducted involved the use of metallic strips as reinforcing material.
Retaining walls with reinforced earth have been constructed around the world since
Vidal began his work. The first reinforced-earth retaining wall with metal strips as reinforcement in the United States was constructed in 1972 in southern California.
The beneficial effects of soil reinforcement derive from (a) the soil’s increased tensile
strength and (b) the shear resistance developed from the friction at the soil-reinforcement
interfaces. Such reinforcement is comparable to that of concrete structures. Currently, most
reinforced-earth design is done with free-draining granular soil only. Thus, the effect of
pore water development in cohesive soils, which, in turn, reduces the shear strength of the
soil, is avoided.
406 Chapter 8: Retaining Walls
8.12
Considerations in Soil Reinforcement
Metal Strips
In most instances, galvanized steel strips are used as reinforcement in soil. However, galvanized steel is subject to corrosion. The rate of corrosion depends on several environmental
factors. Binquet and Lee (1975) suggested that the average rate of corrosion of galvanized
steel strips varies between 0.025 and 0.050 mm>yr. So, in the actual design of reinforcement, allowance must be made for the rate of corrosion. Thus,
tc 5 tdesign 1 r (life span of structure)
where
tc 5 actual thickness of reinforcing strips to be used in construction
tdesign 5 thickness of strips determined from design calculations
r 5 rate of corrosion
Further research needs to be done on corrosion-resistant materials such as fiberglass before
they can be used as reinforcing strips.
Nonbiodegradable Fabrics
Nonbiodegradable fabrics are generally referred to as geotextiles. Since 1970, the use of geotextiles in construction has increased greatly around the world. The fabrics are usually made
from petroleum products—polyester, polyethylene, and polypropylene. They may also be
made from fiberglass. Geotextiles are not prepared from natural fabrics, because they decay
too quickly. Geotextiles may be woven, knitted, or nonwoven.
Woven geotextiles are made of two sets of parallel filaments or strands of yarn systematically interlaced to form a planar structure. Knitted geotextiles are formed by interlocking a series of loops of one or more filaments or strands of yarn to form a planar
structure. Nonwoven geotextiles are formed from filaments or short fibers arranged in an
oriented or random pattern in a planar structure. These filaments or short fibers are
arranged into a loose web in the beginning and then are bonded by one or a combination
of the following processes:
1. Chemical bonding—by glue, rubber, latex, a cellulose derivative, or the like
2. Thermal bonding—by heat for partial melting of filaments
3. Mechanical bonding—by needle punching
Needle-punched nonwoven geotextiles are thick and have high in-plane permeability.
Geotextiles have four primary uses in foundation engineering:
1. Drainage: The fabrics can rapidly channel water from soil to various outlets,
thereby providing a higher soil shear strength and hence stability.
2. Filtration: When placed between two soil layers, one coarse grained and the other fine
grained, the fabric allows free seepage of water from one layer to the other. However,
it protects the fine-grained soil from being washed into the coarse-grained soil.
3. Separation: Geotextiles help keep various soil layers separate after construction and
during the projected service period of the structure. For example, in the construction
of highways, a clayey subgrade can be kept separate from a granular base course.
4. Reinforcement: The tensile strength of geofabrics increases the load-bearing capacity of the soil.
8.12 Considerations in Soil Reinforcement
407
Geogrids
Geogrids are high-modulus polymer materials, such as polypropylene and polyethylene,
and are prepared by tensile drawing. Netlon, Ltd., of the United Kingdom was the first producer of geogrids. In 1982, the Tensar Corporation, presently Tensar International
Corporation, introduced geogrids into the United States.
Commercially available geogrids may be categorized by manufacturing process,
principally: extruded, woven, and welded. Extruded geogrids are formed using a thick
sheet of polyethylene or polypropylene that is punched and drawn to create apertures and
to enhance engineering properties of the resulting ribs and nodes. Woven geogrids are
manufactured by grouping polymeric—usually polyester and polypropylene—and weaving them into a mesh pattern that is then coated with a polymeric lacquer. Welded geogrids
are manufactured by fusing junctions of polymeric strips. Extruded geogrids have shown
good performance when compared to other types for pavement reinforcement applications.
Geogrids generally are of two types: (a) uniaxial and (b) biaxial. Figures 8.22a and
b shows these two types of geogrids, which are produced by Tensar International
Corporation.
Uniaxial TENSAR grids are manufactured by stretching a punched sheet of extruded
high-density polyethylene in one direction under carefully controlled conditions. The
process aligns the polymer’s long-chain molecules in the direction of draw and results in a
product with high one-directional tensile strength and a high modulus. Biaxial TENSAR
grids are manufactured by stretching the punched sheet of polypropylene in two orthogonal
directions. This process results in a product with high tensile strength and a high modulus in
two perpendicular directions. The resulting grid apertures are either square or rectangular.
The commercial geogrids currently available for soil reinforcement have nominal rib
thicknesses of about 0.5 to 1.5 mm (0.02 to 0.06 in.) and junctions of about 2.5 to 5 mm (0.1 to
0.2 in.). The grids used for soil reinforcement usually have openings or apertures that are rectangular or elliptical. The dimensions of the apertures vary from about 25 to 150 mm
(1 to 6 in.). Geogrids are manufactured so that the open areas of the grids are greater than
50% of the total area. They develop reinforcing strength at low strain levels, such as 2%
(Carroll, 1988). Table 8.4 gives some properties of the TENSAR biaxial geogrids that are
currently available commercially.
60
Roll
Length
(Longitudinal)
Roll
Length
(Longitudinal)
(b)
Roll
Width
(Transverse)
Roll
Width
(Transverse)
(c)
(a)
Figure 8.22 Geogrid: (a) uniaxial; (b) biaxial; (c) with triangular apertures
(Courtesy of Tensar International Corporation)
408 Chapter 8: Retaining Walls
Table 8.4 Properties of TENSAR Biaxial Geogrids
Geogrid
Property
Aperture size
Machine
direction
Cross-machine
direction
Open area
Junction
Thickness
Tensile modulus
Machine
direction
Cross-machine
direction
Material
Polypropylene
Carbon black
BX1000
BX1100
BX1200
25 mm
(nominal)
33 mm
(nominal)
70% (minimum)
25 mm
(nominal)
33 mm
(nominal)
74% (nominal)
25 mm
(nominal)
33 mm
(nominal)
77% (nominal)
2.3 mm
(nominal)
2.8 mm
(nominal)
4.1 mm
(nominal)
182 kN>m
(minimum)
182 kN>m
(minimum)
204 kN>m
(minimum)
292 kN>m
(minimum)
270 kN>m
(minimum)
438 kN>m
(minimum)
97% (minimum)
2% (minimum)
99% (nominal)
1% (nominal)
99% (nominal)
1% (nominal)
The major function of geogrids is reinforcement. They are relatively stiff. The apertures are large enough to allow interlocking with surrounding soil or rock (Figure 8.23)
to perform the function of reinforcement or segregation (or both). Sarsby (1985) investigated the influence of aperture size on the size of soil particles for maximum frictional
efficiency (or efficiency against pullout). According to this study, the highest efficiency
occurs when
BGG ⬎ 3.5D50
Figure 8.23 Geogrid apertures allowing interlocking with surrounding soil
(8.30)
8.13 General Design Considerations
409
where
BGG ⫽ minimum width of the geogrid aperture
D50 ⫽ the particle size through which 50% of the backfill soil passes (i.e., the average particle size)
More recently, geogrids with triangular apertures (Figure 8.22c) have been introduced for construction purposes. TENSAR geogrids with triangular apertures are manufactured from a punched polypropylene sheet, which is then oriented in three
substantially equilateral directions so that the resulting ribs shall have a high degree of
molecular orientation. Table 8.5 gives some properties of TENSAR geogrids with triangular apertures.
Table 8.5 Properties of TENSAR Geogrids with Triangular Apertures
Geogrid
Property
Longitudinal
Diagonal
Transverse
TX 160
Rib pitch, (mm)
Mid-rib depth, (mm)
Mid-rib width, (mm)
Nodal thickness, (mm)
Radial stiffness at low strain,
(kN/m @ 0.5% strain)
Rib pitch, (mm)
Mid-rib depth, (mm)
Mid-rib width, (mm)
Nodal thickness, (mm)
Radial stiffness at low strain,
(kN/m @ 0.5% strain)
40
—
—
40
1.8
1.1
—
1.5
1.3
TX 170
8.13
General
3.1
430
40
—
—
40
2.3
1.2
—
1.8
1.3
4.1
475
General Design Considerations
The general design procedure of any mechanically stabilized retaining wall can be divided
into two parts:
1. Satisfying internal stability requirements
2. Checking the external stability of the wall
The internal stability checks involve determining tension and pullout resistance in the reinforcing elements and ascertaining the integrity of facing elements. The external stability
checks include checks for overturning, sliding, and bearing capacity failure (Figure 8.24).
The sections that follow will discuss the retaining-wall design procedures for use with
metallic strips, geotextiles, and geogrids.
410 Chapter 8: Retaining Walls
(a) Sliding
(b) Overturning
(c) Bearing capacity
(d) Deep-seated stability
Figure 8.24 External stability checks (After Transportation Research Board, 1995) (From
Transportation Research Circular 444: Mechanically Stabilized Earth Walls. Transportation
Research Board, National Research Council, Washington, D.C., 1995, Figure 3, p. 7. Reproduced
with permission of the Transportation Research Board.)
8.14
Retaining Walls with Metallic Strip Reinforcement
Reinforced-earth walls are flexible walls. Their main components are
1. Backfill, which is granular soil
2. Reinforcing strips, which are thin, wide strips placed at regular intervals, and
3. A cover or skin, on the front face of the wall
Figure 8.25 is a diagram of a reinforced-earth retaining wall. Note that, at any depth,
the reinforcing strips or ties are placed with a horizontal spacing of SH center to center;
the vertical spacing of the strips or ties is SV center to center. The skin can be constructed
with sections of relatively flexible thin material. Lee et al. (1973) showed that, with a conservative design, a 5 mm-thick galvanized steel skin would be enough to hold a wall about
14 to 15 m high. In most cases, precast concrete slabs can also be used as skin. The
slabs are grooved to fit into each other so that soil cannot flow out between the joints.
When metal skins are used, they are bolted together, and reinforcing strips are placed
between the skins.
Figures 8.26 and 8.27 show a reinforced-earth retaining wall under construction; its
skin (facing) is a precast concrete slab. Figure 8.28 shows a metallic reinforcement tie
attached to the concrete slab.
The simplest and most common method for the design of ties is the Rankine method.
We discuss this procedure next.
8.14 Retaining Walls with Metallic Strip Reinforcement
411
Tie
Skin
SH
SV
Figure 8.25 Reinforced-earth retaining wall
Figure 8.26 Reinforced-earth
retaining wall (with metallic
strip) under construction
(Courtesy of Braja M. Das,
Henderson, NV)
412 Chapter 8: Retaining Walls
Figure 8.27 Another view of the retaining wall shown in Figure 8.26 (Courtesy of Braja M. Das,
Henderson, NV)
Figure 8.28 Metallic strip attachment to the precast concrete slab used as the skin (Courtesy of
Braja M. Das, Henderson, NV)
8.14 Retaining Walls with Metallic Strip Reinforcement
413
Calculation of Active Horizontal and Vertical Pressure
Figure 8.29 shows a retaining wall with a granular backfill having a unit weight of g1 and a
friction angle of f1r . Below the base of the retaining wall, the in situ soil has been excavated
and recompacted, with granular soil used as backfill. Below the backfill, the in situ soil has a
unit weight of g2 , friction angle of f2r , and cohesion of c2r . A surcharge having an intensity of
q per unit area lies atop the retaining wall, which has reinforcement ties at depths
z 5 0, SV , 2SV , c, NSV . The height of the wall is NSV 5 H.
According to the Rankine active pressure theory (Section 7.3)
sar 5 sor Ka 2 2cr"Ka
where sar 5 Rankine active pressure at any depth z.
For dry granular soils with no surcharge at the top, cr 5 0, sor 5 g1z, and Ka 5 tan2
(45 2 f1r >2). Thus,
sa(1)
r 5 g1zKa
When a surcharge is added at the top, as shown in Figure 8.29,
b
a
q/unit area
C
A
45 ␾1/2
SV
Sand
␥1
␾1
SV
z
lr
(a)
le
SV
H
SV
SV
SV
z NSV
B
In situ soil
␥2; ␾2; c2
(b)
␴a(1) ␴a(2)
␴a
Ka␥1z
Figure 8.29 Analysis of a reinforced-earth retaining wall
(8.31)
414 Chapter 8: Retaining Walls
sor 5 so(1)
r
c
5 g1z
Due to
soil only
1 so(2)
r
c
Due to the
surcharge
(8.32)
The magnitude of so(2)
r can be calculated by using the 2:1 method of stress distribution described in Eq. (5.14) and Figure 5.5. The 2:1 method of stress distribution is shown
in Figure 8.30a. According to Laba and Kennedy (1986),
so(2)
r 5
qar
ar 1 z
(for z # 2br)
(8.33)
and
so(2)
r 5
qar
z
ar 1 1 br
2
(for z . 2br)
(8.34)
Also, when a surcharge is added at the top, the lateral pressure at any depth is
r
sar 5 sa(1)
c
5 Kag1z
Due to
soil only
b
1 sa(2)
r
c
Due to the
surcharge
(8.35)
b
a
a
q/unit area
q/unit area
␤
z
z
H
2
2
1
1
Sand
␥1; ␾1
H
␴o(2)
Reinforcement
strip
(a)
␣
␴a(2)
Sand
␥1; ␾1
Reinforcement
strip
(b)
Figure 8.30 (a) Notation for the relationship of so(2)
r in Eqs. (8.33) and (8.34); (b) notation for
the relationship of sa(2)
r in Eqs. (8.36) and (8.37)
415
8.14 Retaining Walls with Metallic Strip Reinforcement
r may be expressed (see Figure 8.30b) as
According to Laba and Kennedy (1986), sa(2)
2q
(b 2 sin b cos 2a) R
p
sa(2)
r 5 MB
c
(in radians)
(8.36)
where
M 5 1.4 2
0.4br
$1
0.14H
(8.37)
The net active (lateral) pressure distribution on the retaining wall calculated by using Eqs.
(8.35), (8.36), and (8.37) is shown in Figure 8.29b.
Tie Force
The tie force per unit length of the wall developed at any depth z (see Figure 8.29) is
T 5 active earth pressure at depth z
3 area of the wall to be supported by the tie
5 (sar ) (SVSH )
(8.38)
Factor of Safety against Tie Failure
The reinforcement ties at each level, and thus the walls, could fail by either (a) tie breaking or (b) tie pullout.
The factor of safety against tie breaking may be determined as
FS(B) 5
5
yield or breaking strength of each tie
maximum force in any tie
wtfy
(8.39)
sar SVSH
where
w 5 width of each tie
t 5 thickness of each tie
fy 5 yield or breaking strength of the tie material
A factor of safety of about 2.5 to 3 is generally recommended for ties at all levels.
Reinforcing ties at any depth z will fail by pullout if the frictional resistance
developed along the surfaces of the ties is less than the force to which the ties are being
subjected. The effective length of the ties along which frictional resistance is developed
416 Chapter 8: Retaining Walls
may be conservatively taken as the length that extends beyond the limits of the Rankine
active failure zone, which is the zone ABC in Figure 8.29. Line BC makes an angle of
45 1 f1r >2 with the horizontal. Now, the maximum friction force that can be realized
for a tie at depth z is
FR 5 2lewsor tan fmr
(8.40)
where
le 5 effective length
sor 5 effective vertical pressure at a depth z
fmr 5 soil–tie friction angle
Thus, the factor of safety against tie pullout at any depth z is
FS(P) 5
FR
T
(8.41)
Substituting Eqs. (8.38) and (8.40) into Eq. (8.41) yields
FS(P) 5
2lewsor tan fmr
sar SVSH
(8.42)
Total Length of Tie
The total length of ties at any depth is
L 5 lr 1 le
(8.43)
where
lr 5 length within the Rankine failure zone
le 5 effective length
For a given FS(P) from Eq. (8.42),
le 5
FS(P)sar SVSH
2wsor tan fmr
(8.44)
Again, at any depth z,
lr 5
(H 2 z)
f1r
tan ¢45 1 ≤
2
(8.45)
So, combining Eqs. (8.43), (8.44), and (8.45) gives
L5
(H 2 z)
f1r
tan ¢ 45 1 ≤
2
1
FS(P)sar SVSH
2wsor tan fmr
(8.46)
417
8.15 Step-by-Step-Design Procedure Using Metallic Strip Reinforcement
8.15
Step-by-Step-Design Procedure Using
Metallic Strip Reinforcement
Following is a step-by-step procedure for the design of reinforced-earth retaining walls.
General
Step 1. Determine the height of the wall, H, and the properties of the granular
backfill material, such as the unit weight (g1 ) and the angle of friction
(f1r ).
Step 2. Obtain the soil–tie friction angle, fmr , and the required value of FS(B) and
FS(P) .
Internal Stability
Step 3. Assume values for horizontal and vertical tie spacing. Also, assume the
width of reinforcing strip, w, to be used.
Step 4. Calculate sar from Eqs. (8.35), (8.36), and (8.37).
Step 5. Calculate the tie forces at various levels from Eq. (8.38).
Step 6. For the known values of FS(B) , calculate the thickness of ties, t, required to
resist the tie breakout:
T 5 sar SVSH 5
wtfy
FS(B)
or
t5
(sar SVSH ) 3FS(B) 4
wfy
(8.47)
The convention is to keep the magnitude of t the same at all levels, so sar
in Eq. (8.47) should equal sa(max)
r
.
Step 7. For the known values of fmr and FS(P) , determine the length L of the ties at
various levels from Eq. (8.46).
Step 8. The magnitudes of SV , SH , t, w, and L may be changed to obtain the most
economical design.
External Stability
Step 9. Check for overturning, using Figure 8.31 as a guide. Taking the moment
about B yields the overturning moment for the unit length of the wall:
Mo 5 Pazr
(8.48)
Here,
H
Pa 5 active force 5 3 sar dz
0
The resisting moment per unit length of the wall is
418 Chapter 8: Retaining Walls
b
a
q/unit area
A
I
L L1
x1
z
W1
Sand
␥1; ␾1
F
E
G
Pa
L L2
H
x2
z
W2
B Sand ␥ ; ␾
1 1
D
Figure 8.31 Stability check for the
retaining wall
In situ soil
␥2; ␾2; c2
ar
MR 5 W1x1 1 W2x2 1 c 1 qar ¢ br 1 ≤
2
(8.49)
where
W1 5 (area AFEGI) (1) (g1 )
W2 5 (area FBDE) (1) (g1 )
.(
So,
FS(overturning) 5
MR
Mo
5
ar
W1x1 1 W2x2 1 c1 qar ¢ br 1 ≤
2
(8.50)
H
¢ 3 sar dz ≤ zr
0
Step 10. The check for sliding can be done by using Eq. (8.11), or
FS(sliding) 5
where k < 23 .
(W1 1 W2 1 c 1 qar) 3 tan (kf1r ) 4
Pa
(8.51)
8.15 Step-by-Step-Design Procedure Using Metallic Strip Reinforcement
419
Step 11. Check for ultimate bearing capacity failure, which can be given as
qu 5 c2r Nc 1 12g2L2Ng
(8.52)
The bearing capacity factors Nc and Ng correspond to the soil friction angle
f2r . (See Table 3.3.)
From Eq. 8.32, the vertical stress at z 5 H is
so(H)
r
5 g1H 1 so(2)
r
(8.53)
So the factor of safety against bearing capacity failure is
FS(bearing capacity) 5
qult
so(H)
r
(8.54)
Generally, minimum values of FS(overturning) 5 3, FS(sliding) 5 3, and
FS(bearing capacity failure) 5 3 to 5 are recommended.
Example 8.5
A 10 m high retaining wall with galvanized steel-strip reinforcement in a granular backfill has to be constructed. Referring to Figure 8.29, given:
Granular backfill:
f1r 5 36°
g1 5 16.5 kN>m3
Foundation soil:
f2r 5 28°
g2 5 17.3 kN>m3
c2r 5 50 kN>m2
Galvanized steel reinforcement:
Width of strip,
w 5 75 mm
SV 5 0.6 m center-to-center
SH 5 1 m center-to-center
fy 5 240,00 kN>m2
fmr 5 20°
Required FS(B) 5 3
Required
FS(P) 5 3
Check for the external and internal stability. Assume the corrosion rate of the galvanized
steel to be 0.025 mm> year and the life span of the structure to be 50 years.
Solution
Internal Stability Check
r
SVSH
Tie thickness: Maximum tie force, Tmax 5 sa(max)
420 Chapter 8: Retaining Walls
sa(max) 5 g1HKa 5 g1H tan2 a45 2
f1r
b
2
so
Tmax 5 g1H tan2 a45 2
f1r
bS S
2 V H
From Eq. (8.47), for tie break,
t5
(sar SVSH ) 3FS(B) 4
wfy
5
f1r
bS S d FS(B)
2 V H
wfy
cg1H tan2 a45 2
or
t5
c (16.5) (10) tan2 a45 2
36
b (0.6) (1) d (3)
2
5 0.00428m 5 4.28 mm
(0.075 m) (240,000 kN>m2 )
If the rate of corrosion is 0.025 mm> yr and the life span of the structure is 50 yr, then the
actual thickness, t, of the ties will be
t 5 4.28 1 (0.025) (50) 5 5.53 mm
So a tie thickness of 6 mm would be enough.
Tie length: Refer to Eq. (8.46). For this case, sar 5 g1zKa and sor 5 g1z, so
FS(P)g1zKaSVSH
(H 2 z)
1
f1r
2wg1z tanfmr
tana45 1
b
2
L5
Now the following table can be prepared. (Note: FS(P) 5 3, H 5 10 m, w 5 75 mm,
and fmr 5 20°.)
z (m)
Tie length L (m)
[Eq. (8.46)]
2
4
6
8
10
12.65
11.63
10.61
9.59
8.57
So use a tie length of L 5 13 m.
External Stability Check
Check for overturning: Refer to Figure 8.32. For this case, using Eq. (8.50)
FS(overturning) 5
W1x1
H
c 3 sar dz d zr
0
8.15 Step-by-Step-Design Procedure Using Metallic Strip Reinforcement
421
␥1 16.5 kN/m3
␾1 36°
6.5 m
10 m
W1
L 13 m
␥2 17.3 kN/m3
␾2 28°
c2 50 kN/m2
Figure 8.32 Retaining wall with galvanized
steel-strip reinforcement in the backfill
W1 5 g1HL 5 (16.5) (10) (13) 5 2145 kN>m
x1 5 6.5 m
H
Pa 5 3 sar dz 5 12g1KaH 2 5 ( 12 ) (16.5) (0.26) (10) 2 5 214.5 kN>m
0
zr 5
FS(overturning) 5
10
5 3.33 m
3
(2145) (6.5)
5 19.52 + 3—OK
(214.5) (3.33)
Check for sliding: From Eq. (8.51)
FS(sliding)
W1 tan(kf1r )
5
5
Pa
2
2145 tan c a b (36) d
3
5 4.45 + 3—OK
214.5
Check for bearing capacity: For f2r 5 28°, Nc 5 25.8, Ng 5 16.78 (Table 3.3). From
Eq. (8.52),
qult 5 c2r Nc 1 12g2L Ng
qult 5 (50) (25.8) 1 ( 12 ) (17.3) (13) (16.72) 5 3170.16 kN>m2
From Eq. (8.53),
so(H)
r
5 g1H 5 (16.5) (10) 5 165 kN>m2
FS(bearing capacity) 5
qult
3170.16
5
5 19.2 + 5—OK
so(H)
r
165
■
422 Chapter 8: Retaining Walls
8.16
Retaining Walls with Geotextile Reinforcement
Figure 8.33 shows a retaining wall in which layers of geotextile have been used as reinforcement. As in Figure 8.31, the backfill is a granular soil. In this type of retaining wall,
the facing of the wall is formed by lapping the sheets as shown with a lap length of ll . When
construction is finished, the exposed face of the wall must be covered; otherwise, the geotextile will deteriorate from exposure to ultraviolet light. Bitumen emulsion or Gunite is
sprayed on the wall face. A wire mesh anchored to the geotextile facing may be necessary
to keep the coating on. Figure 8.34 shows the construction of a geotextile-reinforced retaining wall. Figure 8.35 shows a completed geosynthetic-reinforced soil wall. The wall is in
DeBeque Canyon, Colorado. Note the versatility of the facing type. In this case, single-tier
concrete block facing is integrated with a three-tier facing via rock facing.
SV
Geotextile
SV
z
lr
H
45 ␾1/2
In situ soil
␥2; ␾2; c2
ll
SV
Geotextile
le
SV Sand,␥1; ␾1
Geotextile
SV
Geotextile
Geotextile
Figure 8.33 Retaining wall with
geotextile reinforcement
Figure 8.34 Construction of a geotextile-reinforced retaining wall (Courtesy of Jonathan
T. H. Wu, University of Colorado at Denver, Denver, Colorado)
423
8.16 Retaining Walls with Geotextile Reinforcement
Figure 8.35 A completed geotextile-reinforced retaining wall in DeBeque Canyon, Colorado
(Courtesy of Jonathan T. H. Wu, University of Colorado at Denver, Denver, Colorado)
The design of this type of retaining wall is similar to that presented in Section 8.15.
Following is a step-by-step procedure for design based on the recommendations of Bell
et al. (1975) and Koerner (2005):
Internal Stability
Step 1. Determine the active pressure distribution on the wall from the formula
sar 5 Kasor 5 Kag1z
(8.55)
where
Ka 5 Rankine active pressure coefficient 5 tan2 (45 2 f1r >2)
g1 5 unit weight of the granular backfill
f1r 5 friction angle of the granular backfill
Step 2. Select a geotextile fabric with an allowable tensile strength, Tall (lb>ft or
kN>m ).
The allowable tensile strength for retaining wall construction may
be expressed as (Koerner, 2005)
Tall 5
Tult
RFid 3 RFcr 3 RFcbd
where
Tult ⫽ ultimate tensile strength
RFid ⫽ reduction factor for installation damage
RFcr ⫽ reduction factor for creep
RFcbd ⫽ reduction factor for chemical and biological degradation
(8.56)
424 Chapter 8: Retaining Walls
The recommended values of the reduction factor are as follows (Koerner, 2005)
RFid
RFcr
RFcbd
1.1–2.0
2–4
1–1.5
Step 3. Determine the vertical spacing of the layers at any depth z from the formula
SV 5
Tall
Tall
5
sar FS(B)
(g1zKa ) 3FS(B) 4
(8.57)
Note that Eq. (8.57) is similar to Eq. (8.39). The magnitude of FS(B) is generally 1.3 to 1.5.
Step 4. Determine the length of each layer of geotextile from the formula
L 5 lr 1 le
(8.58)
where
lr 5
H2z
(8.59)
f1r
tan ¢45 1 ≤
2
and
le 5
SVsar 3FS(P) 4
(8.60)
2sor tan fFr
in which
sar 5 g1zKa
sor 5 g1z
FS(P) 5 1.3 to 1.5
fFr 5 friction angle at geotextile–soil interface
< 23f1r
Note that Eqs. (8.58), (8.59), and (8.60) are similar to Eqs. (8.43), (8.45),
and (8.44), respectively.
Based on the published results, the assumption of fFr >f1r < 23 is reasonable
and appears to be conservative. Martin et al. (1984) presented the following laboratory test results for fFr >f1r between various types of geotextiles and sand.
Type
Woven—monofilament> concrete sand
Woven—silt film> concrete sand
Woven—silt film> rounded sand
Woven—silt film> silty sand
Nonwoven—melt-bonded> concrete sand
Nonwoven—needle-punched> concrete sand
Nonwoven—needle-punched> rounded sand
Nonwoven—needle-punched> silty sand
fF9 , f19
0.87
0.8
0.86
0.92
0.87
1.0
0.93
0.91
8.16 Retaining Walls with Geotextile Reinforcement
Step 5. Determine the lap length, ll , from
SVsar FS(P)
ll 5
4sor tan fFr
425
(8.61)
The minimum lap length should be 1 m.
External Stability
Step 6. Check the factors of safety against overturning, sliding, and bearing capacity failure as described in Section 8.15 (Steps 9, 10, and 11).
Example 8.6
A geotextile-reinforced retaining wall 5 m high is shown in Figure 8.36. For the granular backfill, g1 5 15.7 kN>m3 and f1r 5 36°. For the geotextile, Tult 5 52.5 kN>m. For
the design of the wall, determine SV , L, and ll . Use RFid 5 1.2, RFcr 5 2.5,
and RFcbd 5 1.25.
Solution
We have
Ka 5 tan2 ¢ 45 2
f1r
≤ 5 0.26
2
Determination of SV
To find SV , we make a few trials. From Eq. (8.57),
Tall
SV 5
(g1zKa ) 3FS(B) 4
2.5 m
SV = 0.5 m
5m
␥1 = 15.7 kN/m3
␾1 = 36°
ll = l m
␥2 = 18 kN/m3
␾2 = 22°
c2 = 28 kN/m2
Figure 8.36 Geotextile-reinforced retaining wall
426 Chapter 8: Retaining Walls
From Eq. (8.56),
Tall 5
Tuef
52.5
5
5 14 kN>m
RFid 3 RFcr 3 RFcbd 1.2 3 2.5 3 1.25
With FS(B) 5 1.5 at z 5 2 m,
SV 5
14
5 1.14 m
(15.7) (2) (0.26) (1.5)
SV 5
14
5 0.57 m
(15.7) (4) (0.26) (1.5)
SV 5
14
5 0.46 m
(15.7) (5) (0.26) (1.5)
At z 5 4 m,
At z 5 5 m,
So, use SV 5 0.5 m for z 5 0 to z 5 5 m (See Figure 8.36.)
Determination of L
From Eqs. (8.58), (8.59), and (8.60),
(H 2 z)
L5
tan ¢ 45 1
f1r
≤
2
1
SVKa 3FS(P) 4
2 tan fFr
For FS(P) 5 1.5, tan fFr 5 tan 3( 23 ) (36)4 5 0.445, and it follows that
L 5 (0.51) (H 2 z) 1 0.438SV
H ⫽ 5 m, SV ⫽ 0.5 m
At z ⫽ 0.5 m: L ⫽ (0.51)(5 ⫺ 0.5) ⫹ (0.438)(0.5) ⫽ 2.514 m
At z ⫽ 2.5 m: L ⫽ (0.51)(5 ⫺ 2.5) ⫹ (0.438)(0.5) ⫽ 1.494 m
So, use L ⴝ 2.5 m throughout.
Determination of ll
From Eq. (8.61),
ll 5
SVsar 3FS(P) 4
4sor tan fFr
sar 5 g1zKa , FS(P) 5 1.5; with sor 5 g1z, fFr 5 23f1r . So
ll 5
SVKa 3FS(P) 4
So, use l l 5 1 m.
5
SV (0.26) (1.5)
5 0.219SV
4 tan 3( 23 ) (36) 4
ll 5 0.219SV 5 (0.219) (0.5) 5 0.11 m # 1 m
4 tan fFr
■
8.16 Retaining Walls with Geotextile Reinforcement
427
Example 8.7
Consider the results of the internal stability check given in Example 8.6. For the
geotextile-reinforced retaining wall, calculate the factor of safety against overturning, sliding, and bearing capacity failure.
Solution
Refer to Figure 8.37.
Factor of Safety Against Overturning
From Eq. (8.50), FS (overturning) 5
W1x1
(Pa ) a
H
b
3
W1 ⫽ (5)(2.5)(15.7) ⫽ 196.25 kN/m
x1 5
2.5
5 1.25 m
2
1
1
Pa 5 gH 2Ka 5 a b (15.7) (5) 2 (0.26) 5 51.03 kN>m
2
2
Hence,
FS (overturning) 5
(196.25) (1.25)
5 2.88 , 3
51.03(5>3)
(increase length of geotextile layers to 3 m)
2.5 m
SV = 0.5 m
x1
5m
W1
␥1 = 15.7 kN/m3
␾1 = 36°
ll = 1 m
␥2 = 18 kN/m3
␾2 = 22°
c2 = 28 kN/m2
Figure 8.37 Stability check
428 Chapter 8: Retaining Walls
Factor of Safety Against Sliding
From Eq. (8.51),
FS (sliding)
2
2
W1tana f1rb
(196.25) ctana 3 36b d
3
3
5
5
5 1.71 + 1.5 2 O.K.
Pa
51.03
Factor of Safety Against Bearing Capacity Failure
1
From Eq. (8.52), qu 5 c2r Nc 1 g2 L2 Ng
2
Given: ␥2 ⫽ 18 kN/m3, L2 ⫽ 2.5 m, c2⬘ ⫽ 28 kN/m2, and ␾2⬘ ⫽ 22°. From Table 3.3,
Nc ⫽ 16.88, and N␥ ⫽ 7.13.
1
qu 5 (28) (16.88) 1 a b (18) (2.5) (7.13) < 633 kN>m2
2
From Eq. (8.54),
FS(bearing capacity) 5
8.17
qu
so9(H)
5
633
633
5
5 8.06 + 3 2 O.K.
g1H
(15.7) (5)
■
Retaining Walls with Geogrid Reinforcement—General
Geogrids can also be used as reinforcement in granular backfill for the construction of retaining
walls. Figure 8.38 shows typical schematic diagrams of retaining walls with geogrid reinforcement. Figure 8.39 shows some photographs of geogrid-reinforced retaining walls in the field.
Relatively few field measurements are available for lateral earth pressure on retaining
walls constructed with geogrid reinforcement. Figure 8.40 shows a comparison of measured
and design lateral pressures (Berg et al., 1986) for two retaining walls constructed with precast panel facing. The figure indicates that the measured earth pressures were substantially
smaller than those calculated for the Rankine active case.
8.18
Design Procedure for Geogrid-Reinforced
Retaining Wall
Figure 8.41 shows a schematic diagram of a concrete panel-faced wall with a granular
backfill reinforced with layers of geogrid. The design process of the wall is essentially similar to that with geotextile reinforcement of the backfill given in Section 8.16. The following is a brief step-by-step procedure.
Internal Stability
Step 1. Determine the active pressure at any depth z as [similar to Eq. (8.55)]:
␴a⬘ ⫽ Ka␥1z
(8.62)
where
Ka ⫽ Rankine active pressure coefficient ⫽ tan2 a45 2
f91
b
2
8.18 Design Procedure fore Geogrid-Reinforced Retaining Wall
429
Geogrids – biaxial
Geogrids – uniaxial
(a)
Gabion facing
Geogrids
(b)
Precast
concrete
panel
Pinned connection
Geogrids
Leveling pad
(c)
Figure 8.38 Typical schematic diagrams of retaining walls with geogrid reinforcement:
(a) geogrid wraparound wall; (b) wall with gabion facing; (c) concrete panel-faced wall (After The
Tensar Corporation, 1986)
430 Chapter 8: Retaining Walls
(b)
(a)
(c)
0
Figure 8.39 (a) HDPE geogridreinforced wall with precast concrete panel facing under
construction; (b) Mechanical
splice between two pieces of
geogrid in the working direction;
(c) Segmented concrete-block
faced wall reinforced with uniaxial geogrid (Courtesy of Tensar
International Corporation,
Atlanta, Georgia)
Lateral pressure, ␴a (kN/m2)
10
20
30
40
0
Wall at Tuscon, Arizona, H 4.6 m
1
2
3
Wall at Lithonia, Georgia, H 6 m
Measured pressure
Rankine active pressure
4
5
Height of fill above load cell (m)
Figure 8.40 Comparison of theoretical
and measured lateral pressures in geogrid
reinforced retaining walls (Based on Berg
et al., 1986)
8.18 Design Procedure fore Geogrid-Reinforced Retaining Wall
431
W1
z
Granular
backfill
SV
L1
␥1
␾1
H
W2
L2
Leveling pad
Foundation soil
␥2, ␾2, c2
Figure 8.41 Design of geogrid-reinforced retaining wall
Step 2. Select a geogrid with allowable tensile strength, Tall [similar to Eq. (8.56)]
(Koerner, 2005):
Tult
Tall 5
(8.63)
RFid 3 RFcr 3 RFcbd
where
RFid ⫽ reduction factor for installation damage (1.1 to 1.4)
RFcr ⫽ reduction factor for creep (2.0 to 3.0)
RFcbd ⫽ reduction factor for chemical and biological degradation (1.1 to 1.5).
Step 3. Obtain the vertical spacing of the geogrid layers, SV, as
SV 5
TallCr
sra FS(B)
(8.64)
where Cr ⫽ coverage ratio for geogrid.
The coverage ratio is the fractional plan area at any particular elevation
that is actually occupied by geogrid. For example, if there is a 0.3 m (1 ft) wide
space between each 1.2 m (4 ft) wide piece of geogrid, the coverage ratio is
Cr 5
1.2 m
5 0.8
1.2 m 1 0.3 m
Step 4. Calculate the length of each layer of geogrid at a depth z as [Eq. (8.58)]
L ⫽ lr ⫹ le
lr 5
H2z
tan2 a45 2
fr1
b
2
(8.65)
432 Chapter 8: Retaining Walls
For determination of le [similar to Eq. (8.60)],
FS(P) 5
resistance to pullout at a given normal effective stress
pullout force
5
(2) (le ) (Cis0r tan f1r) (Cr )
SVsar
5
(2) (le ) (Ci tan f1r) (Cr )
SVKa
(8.66)
where Ci ⫽ interaction coefficient or
le 5
SVKa FS(P)
(8.67)
2CrCi tan fr1
Thus, at a given depth z, the total length, L, of the geogrid layer is
L 5 lr 1 le 5
SVKa FS(P)
H2z
1
2CrCi tan fr1
f1r
tana45 1 b
2
(8.68)
The interaction coefficient, Ci, can be determined experimentally in the laboratory.
The following is an approximate range for Ci for various types of backfill.
Gravel, sandy gravel
Well graded sand, gravelly sand
Fine sand, silty sand
0.75–0.8
0.7–0.75
0.55–0.6
External Stability
Check the factors of safety against overturning, sliding, and bearing capacity failure as
described in Section 8.15 (Steps 9, 10, and 11).
Example 8.8
Consider a geogrid-reinforced retaining wall. Referring to Figure 8.41, given: H ⫽ 6 m,
␥1 ⫽ 16.5 kN/m3, ␾1⬘ ⫽ 35°, Tall ⫽ 45 kN/m, FS(B) ⫽ 1.5, FS(P) ⫽ 1.5, Cr ⫽ 0.8, and Ci
⫽ 0.75. For the design of the wall, determine SV and L.
Solution
Ka 5 tan2 a45 2
fr1
35
b 5 tan2 a45 2 b 5 0.27
2
2
Determination of SV
From Eq. (8.64),
Sv 5
TallCr
s9a
FS(B)
5
TallCr
(45) (0.8)
5.39
5
5
z
gzKa FS(B)
(16.5) (z) (0.27) (1.5)
Problems
433
5.39
5 2.7 m
2
5.39
At z ⴝ 4 m: Sv 5
5 1.35 m
4
5.39
At z ⴝ 5 m: Sv 5
5 1.08 m
5
Use SV ⬇ 1 m
At z ⴝ 2 m: Sv 5
Determination of L
From Eq. (8.68),
L5
SVKa FS(P)
(1 m) (0.27) (1.5)
62z
H2z
1
5
1
2CrCi tanf1r
(2) (0.8) (0.75) (tan 35°)
f1r
35
tana45 1 b
tana45 1 b
2
2
At z ⴝ 1 m: L ⴝ 0.52(6 ⫺ 1) ⫹ 0.482 ⴝ 3.08 m ⬇ 3.1 m
At z ⴝ 3 m: L ⴝ 0.52(6 ⫺ 3) ⫹ 0.482 ⴝ 2.04 m ⬇ 2.1 m
At z ⴝ 5 m: L ⴝ 0.52(6 ⫺ 5) ⫹ 0.482 ⴝ 1.0 m
So, use L ⴝ 3 m for z ⴝ 0 to 6 m.
■
Problems
In Problems 8.1 through 8.4, use gconcrete ⫽ 23.58 kN> m3. Also, in Eq. (8.11), use k1 ⫽ k2 ⫽
2> 3 and Pp ⫽ 0.
8.1
For the cantilever retaining wall shown in Figure P8.1, let the following data be given:
Wall dimensions:
Soil properties:
8.2
H ⫽ 8 m, x1 ⫽ 0.4 m, x2 ⫽ 0.6 m, x3 ⫽ 1.5 m, x4 ⫽ 3.5 m,
x5 ⫽ 0.96 m, D ⫽ 1.75 m, a ⫽ 10°
g1 ⫽ 16.5 kN> m3, f1r ⫽ 32°, g2 ⫽ 17.6 kN> m3, f2r 5 28°, cr2 ⫽
30 kN> m2
Calculate the factor of safety with respect to overturning, sliding, and bearing capacity.
Repeat Problem 8.1 with the following:
Wall dimensions:
H ⫽ 6.5 m, x1 ⫽ 0.3 m, x2 ⫽ 0.6 m, x3 ⫽ 0.8 m, x4 ⫽ 2 m,
x5 ⫽ 0.8 m, D ⫽ 1.5 m, a ⫽ 0°
g1 ⫽ 18.08 kN/m3, f1r ⫽ 36°, g2 ⫽ 19.65 kN/m3, f2r ⫽ 15°,cr2 ⫽
30 kN> m2
A gravity retaining wall is shown in Figure P8.3. Calculate the factor of safety with
respect to overturning and sliding, given the following data:
Soil properties:
8.3
Wall dimensions:
Soil properties:
H ⫽ 6 m, x1 ⫽ 0.6 m, x2 ⫽ 2 m, x3 ⫽ 2 m, x4 ⫽ 0.5 m, x5 ⫽
0.75 m, x6 ⫽ 0.8 m, D ⫽ 1.5 m
g1 ⫽ 16.5 kN> m3, f1r ⫽ 32°, g2 ⫽ 18 kN/m3, f2r ⫽ 22°, c2r ⫽
40 kN> m2
Use the Rankine active earth pressure in your calculation.
434 Chapter 8: Retaining Walls
␣
x1
␥1
c1 0
␾1
H
D
x5
x2
x3
x4
␥2
␾2
c2
H
␥1
c1 0
␾1
x5
x6
D
x4
x2
x1
x3
␥2
␾2
c2
8.4
Figure P8.1
Figure P8.3
Repeat Problem 8.3 using Coulomb’s active earth pressure in your calculation and
letting d r ⫽ 2> 3 f1r .
8.5 Refer to Figure P8.5 for the design of a gravity retaining wall for earthquake condition Given: kv ⫽ 0 and kh ⫽ 0.3.
a. What should be the weight of the wall for a zero displacement condition? Use a
factor of safety of 2.
b. What should be the weight of the wall for an allowable displacement of 50.8 mm?
References
7m
Sand
8.6
8.7
8.8
435
Sand
␾1 30
␥1 18 kN/m3
␦1 15
␾2 36
␥2 18.5 kN/m3
Figure P8.5
Given: Av ⫽ 0.15 and Aa ⫽ 0.25. Use a factor of safety of 2.
In Figure 8.29a, use the following parameters:
Wall: H ⫽ 8 m
Soil: g1 ⫽ 17 kN> m3, f1r ⫽ 35°
Reinforcement: SV ⫽ 1 m and SH ⫽ 1.5 m
Surcharge: q ⫽ 70 kN> m2, a r ⫽ 1.5 m, and b r ⫽ 2 m
Calculate the vertical stress sor [Eqs. (8.32), (8.33) and (8.34)] at z ⫽ 2 m, 4 m, 6 m
and 8 m.
For the data given in Problem 8.6, calculate the lateral pressure sar at z ⫽ 2 m, 4 m,
6 m and 8 m. Use Eqs. (8.35), (8.36) and (8.37).
A reinforced earth retaining wall (Figure 8.29) is to be 10 m. high. Here,
Backfill: unit weight, g1 ⫽ 16 kN> m3 and soil friction angle, f1r ⫽ 34°
Reinforcement: vertical spacing, SV 5 1 m; horizontal spacing, SH 5 1.25 m;
width of reinforcement 5 120 mm., fy 5 260 MN>m2; fm 5 25°;
factor of safety against tie pullout 5 3; and factor of safety
against tie breaking ⫽ 3
Determine:
a. The required thickness of ties
b. The required maximum length of ties
8.9 In Problem 8.8 assume that the ties at all depths are the length determined in Part b.
For the in situ soil, f2r ⫽ 25°, g2 ⫽ 15.5 kN> m3, c2r ⫽ 30 kN> m2. Calculate the factor
of safety against (a) overturning, (b) sliding, and (c) bearing capacity failure.
8.10 A retaining wall with geotextile reinforcement is 6-m high. For the granular
backfill, g1 ⫽ 15.9 kN> m3 and f1r ⫽ 30°. For the geotextile, Tall ⫽ 16 kN> m. For the
design of the wall, determine SV, L, and ll. Use FS(B) ⫽ FS(P) ⫽ 1.5.
8.11 With the SV, L, and ll determined in Problem 8.10, check the overall stability
(i.e., factor of safety against overturning, sliding, and bearing capacity
failure) of the wall. For the in situ soil, g2 ⫽ 16.8 kN> m3, f2r ⫽ 20°, and c2r ⫽
55 kN> m2.
References
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BELL, J. R., STILLEY, A. N., and VANDRE, B. (1975). “Fabric Retaining Earth Walls,” Proceedings,
Thirteenth Engineering Geology and Soils Engineering Symposium, Moscow, ID.
436 Chapter 8: Retaining Walls
BENTLER, J. G., and LABUZ, J. F. (2006). “Performance of a Cantilever Retaining Wall,” Journal of
Geotechnical and Geoenvironmental Engineering, American Society of Civil Engineers,
Vol. 132, No. 8, pp. 1062–1070.
BERG, R. R., BONAPARTE, R., ANDERSON, R. P., and CHOUERY, V. E. (1986). “Design Construction and
Performance of Two Tensar Geogrid Reinforced Walls,” Proceedings, Third International
Conference on Geotextiles, Vienna, pp. 401– 406.
BINQUET, J., and LEE, K. L. (1975). “Bearing Capacity Analysis of Reinforced Earth Slabs,” Journal
of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 101,
No. GT12, pp. 1257–1276.
CARROLL, R., JR. (1988). “Specifying Geogrids,” Geotechnical Fabric Report, Industrial Fabric Association International, St. Paul, March/April.
CASAGRANDE, L. (1973). “Comments on Conventional Design of Retaining Structure,” Journal of the
Soil Mechanics and Foundations Division, ASCE, Vol. 99, No. SM2, pp. 181–198.
DARBIN, M. (1970). “Reinforced Earth for Construction of Freeways” (in French), Revue Générale
des Routes et Aerodromes, No. 457, September.
DAS, B. M. (1983). Fundamentals of Soil Dynamics, Elsevier, New York.
ELMAN, M. T., and TERRY, C. F. (1988). “Retaining Walls with Sloped Heel,” Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 114, No. GT10, pp. 1194–1199.
KOERNER, R. B. (2005). Design with Geosynthetics, 5th. ed., Prentice Hall, Englewood Cliffs, NJ.
LABA, J. T., and KENNEDY, J. B. (1986). “Reinforced Earth Retaining Wall Analysis and Design,”
Canadian Geotechnical Journal, Vol. 23, No. 3, pp. 317–326.
LEE, K. L., ADAMS, B. D., and VAGNERON, J. J. (1973). “Reinforced Earth Retaining Walls,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers,
Vol. 99, No. SM10, pp. 745 –763.
MARTIN, J. P., KOERNER, R. M., and WHITTY, J. E. (1984). “Experimental Friction Evaluation of Slippage Between Geomembranes, Geotextiles, and Soils,” Proceedings, International Conference on Geomembranes, Denver, pp. 191–196.
RICHARDS, R., and ELMS, D. G. (1979). “Seismic Behavior of Gravity Retaining Walls,” Journal
of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 105,
No. GT4, pp. 449–464.
SARSBY, R. W. (1985). “The Influence of Aperture Size/Particle Size on the Efficiency of Grid Reinforcement,” Proceedings, 2nd Canadian Symposium on Geotextiles and Geomembranes,
Edmonton, pp. 7–12.
SCHLOSSER, F., and LONG, N. (1974). “Recent Results in French Research on Reinforced Earth,”
Journal of the Construction Division, American Society of Civil Engineers, Vol. 100, No.
CO3, pp. 113 –237.
SCHLOSSER, F., and VIDAL, H. (1969). “Reinforced Earth” (in French), Bulletin de Liaison des Laboratoires Routier, Ponts et Chassées, Paris, France, November, pp. 101–144.
TENSAR CORPORATION (1986). Tensar Technical Note. No. TTN:RW1, August.
TERZAGHI, K., and PECK, R. B. (1967). Soil Mechanics in Engineering Practice, Wiley, New York.
TRANSPORTATION RESEARCH BOARD (1995). Transportation Research Circular No. 444, National Research Council, Washington, DC.
VIDAL, H. (1966). “La terre Armée,” Annales de l’Institut Technique du Bâtiment et des Travaux
Publiques, France, July–August, pp. 888–938.
9
9.1
Sheet Pile Walls
Introduction
Connected or semiconnected sheet piles are often used to build continuous walls for waterfront structures that range from small waterfront pleasure boat launching facilities to large
dock facilities. (See Figure 9.1.) In contrast to the construction of other types of retaining
wall, the building of sheet pile walls does not usually require dewatering of the site. Sheet
piles are also used for some temporary structures, such as braced cuts. (See Chapter 10.)
The principles of sheet-pile wall design are discussed in the current chapter.
Several types of sheet pile are commonly used in construction: (a) wooden sheet
piles, (b) precast concrete sheet piles, and (c) steel sheet piles. Aluminum sheet piles are
also marketed.
Wooden sheet piles are used only for temporary, light structures that are above the
water table. The most common types are ordinary wooden planks and Wakefield piles.
The wooden planks are about 50 mm 3 300 mm in cross section and are driven edge
to edge (Figure 9.2a). Wakefield piles are made by nailing three planks together, with
the middle plank offset by 50 to 75 mm (Figure 9.2b). Wooden planks can also be
milled to form tongue-and-groove piles, as shown in Figure 9.2c. Figure 9.2d shows
another type of wooden sheet pile that has precut grooves. Metal splines are driven into
the grooves of the adjacent sheetings to hold them together after they are sunk into the
ground.
Water
table
Sheet
pile
Water table
Land side
Dredge line
Figure 9.1 Example of waterfront sheet-pile wall
437
438 Chapter 9: Sheet Pile Walls
Precast Concrete Sheet Pile
Wooden Sheet Piles
150250 mm
(a) Planks
Concrete
grout
Section
500-800 mm
(b) Wakefield piles
Reinforcement
(c) Tongue-and-groove piles
Elevation
(e)
(d) Splined piles
(not to scale)
Figure 9.2 Various types of wooden and concrete sheet pile
Precast concrete sheet piles are heavy and are designed with reinforcements to
withstand the permanent stresses to which the structure will be subjected after construction and also to handle the stresses produced during construction. In cross section,
these piles are about 500 to 800 mm wide and 150 to 250 mm thick. Figure 9.2e is
a schematic diagram of the elevation and the cross section of a reinforced concrete
sheet pile.
Steel sheet piles in the United States are about 10 to 13 mm thick. European
sections may be thinner and wider. Sheet-pile sections may be Z, deep arch, low arch, or
straight web sections. The interlocks of the sheet-pile sections are shaped like a thumband-finger or ball-and-socket joint for watertight connections. Figure 9.3a is a schematic
diagram of the thumb-and-finger type of interlocking for straight web sections. The balland-socket type of interlocking for Z section piles is shown in Figure 9.3b. Figure 9.4
shows a sheet pile wall. Table 9.1 lists the properties of the steel sheet pile sections
produced by the Bethlehem Steel Corporation. The allowable design flexural stress for
the steel sheet piles is as follows:
Type of steel
Allowable stress
ASTM A-328
ASTM A-572
ASTM A-690
170 MN>m2
210 MN>m2
210 MN>m2
Steel sheet piles are convenient to use because of their resistance to the high driving stress
that is developed when they are being driven into hard soils. Steel sheet piles are also lightweight and reusable.
9.1 Introduction
439
(a)
Figure 9.3 (a) Thumb-and-finger type sheet pile
connection; (b) ball-and-socket type sheet-pile
connection
(b)
(c)
Figure 9.4 A steel sheet pile wall (Courtesy of N. Sivakugan, James Cook University, Australia )
Table 9.1 Properties of Some Sheet-Pile Sections Produced by Bethlehem Steel Corporation
Section modulus
Moment of inertia
m ,m
of wall
m4 , m
of wall
3
Section
designation Sketch of section
326.4 3 1025
PZ-40
670.5 3 1026
12.7 mm
409 mm
15.2 mm
Driving distance
500 mm
(Continued)
440 Chapter 9: Sheet Pile Walls
Table 9.1 (Continued)
Section
designation Sketch of section
PZ-35
Section modulus
Moment of inertia
m3 , m
of wall
m4 , m
of wall
260.5 3 1025
493.4 3 1026
162.3 3 1025
251.5 3 1026
97 3 1025
115.2 3 1026
10.8 3 1025
4.41 3 1026
12.8 3 1025
5.63 3 1026
12.7 mm
379 mm
15.2 mm
Driving distance
575 mm
PZ-27
9.53 mm
304.8 mm
9.53 mm
Driving distance
457.2 mm
PZ-22
9.53 mm
228.6 mm
9.53 mm
Driving distance
PSA-31
12.7 mm
Driving distance
PSA-23
558.8 mm
500 mm
9.53 mm
Driving distance
406.4 mm
9.2 Construction Methods
9.2
441
Construction Methods
Sheet pile walls may be divided into two basic categories: (a) cantilever and (b) anchored.
In the construction of sheet pile walls, the sheet pile may be driven into the ground
and then the backfill placed on the land side, or the sheet pile may first be driven into the
ground and the soil in front of the sheet pile dredged. In either case, the soil used for backfill behind the sheet pile wall is usually granular. The soil below the dredge line may be
sandy or clayey. The surface of soil on the water side is referred to as the mud line or
dredge line.
Thus, construction methods generally can be divided into two categories (Tsinker,
1983):
1. Backfilled structure
2. Dredged structure
The sequence of construction for a backfilled structure is as follows (see Figure 9.5):
Step 1.
Step 2.
Step 3.
Step 4.
Dredge the in situ soil in front and back of the proposed structure.
Drive the sheet piles.
Backfill up to the level of the anchor, and place the anchor system.
Backfill up to the top of the wall.
For a cantilever type of wall, only Steps 1, 2, and 4 apply. The sequence of construction
for a dredged structure is as follows (see Figure 9.6):
Step 1.
Step 2.
Step 3.
Step 4.
Drive the sheet piles.
Backfill up to the anchor level, and place the anchor system.
Backfill up to the top of the wall.
Dredge the front side of the wall.
With cantilever sheet pile walls, Step 2 is not required.
Original
ground
surface
Dredge
Step 1
Dredge
line
Step 2
Anchor
rod
Backfill
Step 3
Backfill
Step 4
Figure 9.5 Sequence of construction for
a backfilled structure
442 Chapter 9: Sheet Pile Walls
Anchor
rod
Original
ground
surface
Backfill
Step 1
Step 2
Backfill
Dredge
Step 3
9.3
Step 4
Figure 9.6 Sequence of construction
for a dredged structure
Cantilever Sheet Pile Walls
Cantilever sheet pile walls are usually recommended for walls of moderate height—
about 6 m or less, measured above the dredge line. In such walls, the sheet piles act
as a wide cantilever beam above the dredge line. The basic principles for estimating
net lateral pressure distribution on a cantilever sheet-pile wall can be explained with
the aid of Figure 9.7. The figure shows the nature of lateral yielding of a cantilever
wall penetrating a sand layer below the dredge line. The wall rotates about point O
(Figure 9.7a). Because the hydrostatic pressures at any depth from both sides of the
wall will cancel each other, we consider only the effective lateral soil pressures. In
zone A, the lateral pressure is just the active pressure from the land side. In zone B,
because of the nature of yielding of the wall, there will be active pressure from the
land side and passive pressure from the water side. The condition is reversed in zone
C—that is, below the point of rotation, O. The net actual pressure distribution on the
wall is like that shown in Figure 9.7b. However, for design purposes, Figure 9.7c
shows a simplified version.
Sections 9.4 through 9.7 present the mathematical formulation of the analysis of
cantilever sheet pile walls. Note that, in some waterfront structures, the water level may
fluctuate as the result of tidal effects. Care should be taken in determining the water level
that will affect the net pressure diagram.
9.4
Cantilever Sheet Piling Penetrating Sandy Soils
To develop the relationships for the proper depth of embedment of sheet piles driven
into a granular soil, examine Figure 9.8a. The soil retained by the sheet piling above
the dredge line also is sand. The water table is at a depth L1 below the top of the wall.
9.4 Cantilever Sheet Piling Penetrating Sandy Soils
Water
table
Zone A
Active
pressure
Sand
Dredge
line
Passive
pressure
Active
pressure
O
Zone B
Active
pressure
Passive
pressure
Zone C Sand
(a)
(b)
(c)
Figure 9.7 Cantilever sheet pile penetrating sand
A
Water
table
Sand
␥
␾
c = 0
L1
␴1
C
Sand
␥sat
␾
c = 0
L
L2
z
P
␴2
Dredge line
D
Slope:
1 vertical:
(Kp – Ka)␥
horizontal
L3
E
D
F
L4
F
z
F
L5
H
␴3
B
(a)
z
␴4
Mmax
Sand
␥sat
␾
G c = 0
(b)
Figure 9.8 Cantilever sheet pile penetrating sand: (a) variation of net pressure diagram;
(b) variation of moment
443
444 Chapter 9: Sheet Pile Walls
Let the effective angle of friction of the sand be fr. The intensity of the active pressure at a depth z 5 L1 is
s1r 5 gL1Ka
(9.1)
where
Ka 5 Rankine active pressure coefficient 5 tan2 (45 2 fr>2)
g 5 unit weight of soil above the water table
Similarly, the active pressure at a depth z 5 L1 1 L2 (i.e., at the level of the dredge
line) is
s2r 5 (gL1 1 grL2 )Ka
(9.2)
where gr 5 effective unit weight of soil 5 gsat 2 gw.
Note that, at the level of the dredge line, the hydrostatic pressures from both sides of
the wall are the same magnitude and cancel each other.
To determine the net lateral pressure below the dredge line up to the point of rotation, O, as shown in Figure 9.7a, an engineer has to consider the passive pressure acting
from the left side (the water side) toward the right side (the land side) of the wall and also
the active pressure acting from the right side toward the left side of the wall. For such
cases, ignoring the hydrostatic pressure from both sides of the wall, the active pressure at
depth z is
sar 5 3gL1 1 grL2 1 gr(z 2 L1 2 L2 )4Ka
(9.3)
Also, the passive pressure at depth z is
spr 5 gr(z 2 L1 2 L2 )Kp
(9.4)
where Kp 5 Rankine passive pressure coefficient 5 tan2 (45 1 fr>2) .
Combining Eqs. (9.3) and (9.4) yields the net lateral pressure, namely,
sr 5 sar 2 spr 5 (gL1 1 grL2 )Ka 2 gr (z 2 L1 2 L2 ) (Kp 2 Ka )
5 sr2 2 gr(z 2 L) (Kp 2 Ka )
(9.5)
where L 5 L1 1 L2.
The net pressure, sr equals zero at a depth L3 below the dredge line, so
s2r 2 gr(z 2 L) (Kp 2 Ka ) 5 0
or
(z 2 L) 5 L3 5
s2r
gr(Kp 2 Ka )
(9.6)
445
9.4 Cantilever Sheet Piling Penetrating Sandy Soils
Equation (9.6) indicates that the slope of the net pressure distribution line DEF is 1 vertical to (Kp 2 Ka )gr horizontal, so, in the pressure diagram,
HB 5 s3r 5 L4 (Kp 2 Ka )gr
(9.7)
At the bottom of the sheet pile, passive pressure, spr , acts from the right toward the left
side, and active pressure acts from the left toward the right side of the sheet pile, so, at
z 5 L 1 D,
spr 5 (gL1 1 grL2 1 grD)Kp
(9.8)
sar 5 grDKa
(9.9)
At the same depth,
Hence, the net lateral pressure at the bottom of the sheet pile is
spr 2 sar 5 s4r 5 (gL1 1 grL2 )Kp 1 grD(Kp 2 Ka )
5 (gL1 1 grL2 )Kp 1 grL3 (Kp 2 Ka ) 1 grL4 (Kp 2 Ka )
(9.10)
5 s5r 1 grL4 (Kp 2 Ka )
where
s5r 5 (gL1 1 grL2 )Kp 1 grL3 (Kp 2 Ka )
D 5 L3 1 L 4
(9.11)
(9.12)
For the stability of the wall, the principles of statics can now be applied:
S horizontal forces per unit length of wall 5 0
and
S moment of the forces per unit length of wall about point B 5 0
For the summation of the horizontal forces, we have
Area of the pressure diagram ACDE 2 area of EFHB 1 area of FHBG 5 0
or
P 2 12s3r L4 1 12L5 (s3r 1 s4r ) 5 0
(9.13)
where P 5 area of the pressure diagram ACDE.
Summing the moment of all the forces about point B yields
L4
L5
1
1
P(L4 1 z) 2 ¢ L4s3r ≤ ¢ ≤ 1 L5 (s3r 1 s4r ) ¢ ≤ 5 0
2
3
2
3
(9.14)
From Eq. (9.13),
L5 5
s3r L4 2 2P
s3r 1 s4r
(9.15)
446 Chapter 9: Sheet Pile Walls
Combining Eqs. (9.7), (9.10), (9.14), and (9.15) and simplifying them further, we obtain
the following fourth-degree equation in terms of L4 :
L44 1 A 1L34 2 A 2L24 2 A 3L4 2 A 4 5 0
(9.16)
In this equation,
A1 5
s5r
gr(Kp 2 Ka )
(9.17)
A2 5
8P
gr(Kp 2 Ka )
(9.18)
A3 5
A4 5
6P32zgr(Kp 2 Ka ) 1 s5r 4
gr2 (Kp 2 Ka ) 2
P(6zs5r 1 4P)
gr2 (Kp 2 Ka ) 2
(9.19)
(9.20)
Step-by-Step Procedure for Obtaining the Pressure Diagram
Based on the preceding theory, a step-by-step procedure for obtaining the pressure diagram
for a cantilever sheet pile wall penetrating a granular soil is as follows:
Step 1.
Step 2.
Step 3.
Step 4.
Step 5.
Step 6.
Step 7.
Step 8.
Step 9.
Step 10.
Step 11.
Step 12.
Step 13.
Calculate Ka and Kp .
Calculate s1r [Eq. (9.1)] and s2r [Eq. (9.2)]. (Note: L1 and L2 will be given.)
Calculate L3 [Eq. (9.6)].
Calculate P.
Calculate z (i.e., the center of pressure for the area ACDE) by taking the
moment about E.
Calculate s5r [Eq. (9.11)].
Calculate A 1 , A 2 , A 3 , and A 4 [Eqs. (9.17) through (9.20)].
Solve Eq. (9.16) by trial and error to determine L4 .
Calculate s4r [Eq. (9.10)].
Calculate s3r [Eq. (9.7)].
Obtain L5 from Eq. (9.15).
Draw a pressure distribution diagram like the one shown in Figure 9.8a.
Obtain the theoretical depth [see Eq. (9.12)] of penetration as L3 1 L4 .
The actual depth of penetration is increased by about 20 to 30%.
Note that some designers prefer to use a factor of safety on the passive earth pressure coefficient at the beginning. In that case, in Step 1,
Kp(design) 5
Kp
FS
where FS 5 factor of safety (usually between 1.5 and 2).
9.4 Cantilever Sheet Piling Penetrating Sandy Soils
447
For this type of analysis, follow Steps 1 through 12 with the value of Ka 5
tan2 (45 2 fr>2) and Kp(design) (instead of Kp). The actual depth of penetration can now be
determined by adding L3 , obtained from Step 3, and L4 , obtained from Step 8.
Calculation of Maximum Bending Moment
The nature of the variation of the moment diagram for a cantilever sheet pile wall is shown
in Figure 9.8b. The maximum moment will occur between points E and Fr. Obtaining the
maximum moment (Mmax ) per unit length of the wall requires determining the point of zero
shear. For a new axis zr (with origin at point E) for zero shear,
P 5 12 (zr) 2 (Kp 2 Ka )gr
or
zr 5
2P
Å (Kp 2 Ka )gr
(9.21)
Once the point of zero shear force is determined (point Fs in Figure 9.8a), the magnitude of the maximum moment can be obtained as
Mmax 5 P(z 1 zr) 2 312 grzr2 (Kp 2 Ka )4 ( 13 )zr
(9.22)
The necessary profile of the sheet piling is then sized according to the allowable flexural
stress of the sheet pile material, or
S5
Mmax
sall
(9.23)
where
S 5 section modulus of the sheet pile required per unit length of the structure
sall 5 allowable flexural stress of the sheet pile
Example 9.1
Figure 9.9 shows a cantilever sheet pile wall penetrating a granular soil. Here, L1 ⫽ 2 m,
L2 ⫽ 3 m, ␥ ⫽ 15.9 kN/m3, ␥sat ⫽ 19.33 kN/m3, and ␾⬘ ⫽ 32°.
a. What is the theoretical depth of embedment, D?
b. For a 30% increase in D, what should be the total length of the sheet piles?
c. What should be the minimum section modulus of the sheet piles? Use ␴all ⫽
172 MN/m2.
Solution
Part a
Using Figure 9.8a for the pressure distribution diagram, one can now prepare the following table for a step-by-step calculation.
448 Chapter 9: Sheet Pile Walls
L1
Sand
␥
c = 0
␾
L2
Sand
␥sat
c = 0
␾
D
Sand
␥sat
c = 0
␾
Water table
Dredge line
Figure 9.9 Cantilever sheet-pile wall
Quantity
required
Eq.
no.
Ka
—
Kp
—
␴1⬘
9.1
␥L1Ka ⫽ (15.9)(2)(0.307) ⫽ 9.763 kN/m2
␴2⬘
9.2
L3
9.6
(␥L1 ⫹ ␥⬘L2)Ka ⫽ [(15.9)(2) ⫹ (19.33 ⫺ 9.81)(3)](0.307) ⫽ 18.53 kN/m2
s2r
18.53
5
5 0.66 m
gr (Kp 2 Ka )
(19.33 2 9.81) (3.25 2 0.307)
P
—
Equation and calculation
fr
32
b 5 tan2 a45 2 b 5 0.307
2
2
r
f
32
tan2 a45 1 b 5 tan2 a45 1 b 5 3.25
2
2
tan2 a45 2
1
1
1
2 s1r L1 1 s1r L2 1 2 (s2r 2 s1r )L2 1 2 s2r L3
1
5 ( 2 ) (9.763) (2) 1 (9.763) (3) 1 ( 12 ) (18.53
1 ( 12 ) (18.53) (0.66)
2 9.763) (3)
5 9.763 1 29.289 1 13.151 1 6.115 5 58.32 kN>m
z
—
␴5⬘
9.11
A1
9.17
A2
9.18
SME
1
9.763(0.66 1 3 1 23 ) 1 29.289(0.66 1 32 )
5
C
S 5 2.23 m
P
58.32 1 13.151(0.66 1 33 ) 1 6.115(0.66 3 23 )
(␥L1 ⫹ ␥⬘L2)Kp ⫹ ␥⬘L3(Kp ⫺ Ka) ⫽ [(15.9)(2) ⫹ (19.33 ⫺ 9.81)(3)](3.25)
⫹ (19.33 ⫺ 9.81)(0.66)(3.25
⫺ 0.307) ⫽ 214.66 kN/m2
s5r
214.66
5
5 7.66
gr (Kp 2 Ka )
(19.33 2 9.81) (3.25 2 0.307)
(8) (58.32)
8P
5
5 16.65
gr (Kp 2 Ka )
(19.33 2 9.81) (3.25 2 0.307)
9.5 Special Cases for Cantilever Walls Penetrating a Sandy Soil
A3
9.19
6P32zgr (Kp 2 Ka ) 1 s5r 4
5
gr2 (Kp 2 Ka ) 2
(6) (58.32) 3 (2) (2.23) (19.33 2 9.81) (3.25 2 0.307) 1 214.664
(19.33 2 9.81) 2 (3.25 2 0.307) 2
5 151.93
A4
9.20
449
P(6zs5r 1 4P)
2
gr (Kp 2 Ka )
2
5
58.323(6) (2.23) (214.66) 1 (4) (58.32) 4
(19.33 2 9.81) 2 (3.25 2 0.307) 2
5 230.72
L4
9.16
L44 ⫹ A1L43 ⫺ A2L42 ⫺ A3L4 ⫺ A4 ⫽ 0
L44 ⫹ 7.66L43 ⫺ 16.65L42 ⫺ 151.93L4 ⫺ 230.72 ⫽ 0; L4 ⬇ 4.8 m
Thus,
Dtheory ⫽ L3 ⫹ L4 ⫽ 0.66 ⫹ 4.8 ⫽ 5.46 m
Part b
The total length of the sheet piles is
L1 ⫹ L2 ⫹ 1.3(L3 ⫹ L4) ⫽ 2 ⫹ 3 ⫹ 1.3(5.46) ⫽ 12.1 m
Part c
Finally, we have the following table.
Quantity
required
Eq.
no.
z⬘
9.21
Mmax
9.22
Equation and calculation
(2) (58.32)
2P
5
5 2.04 m
Ä (Kp 2 Ka )gr Å (3.25 2 0.307) (19.33 2 9.81)
1
zr
P(z 1 z r ) 2 c grz r2 (Kp 2 Ka ) d 5 (58.32) (2.23 1 2.04)
2
3
1
2.04
2 c a b (19.33 2 9.81) (2.04) 2 (3.25 2 0.307) d
2
3
5 209.39 kN # m>m
S
9.5
9.29
Mmax
209.39 kN # m
5
5 1.217 3 1023 m3>m of wall
s all
172 3 103 kN>m2
■
Special Cases for Cantilever Walls Penetrating
a Sandy Soil
Sheet Pile Wall with the Absence of Water Table
In the absence of the water table, the net pressure diagram on the cantilever sheet-pile wall
will be as shown in Figure 9.10, which is a modified version of Figure 9.8. In this case,
450 Chapter 9: Sheet Pile Walls
Sand
␥
␾
L
P
z
␴2
Sand
␥
␾
L3
D
L4
Figure 9.10 Sheet piling penetrating
a sandy soil in the absence of the
water table
L5
␴3
␴4
s2r 5 gLKa
s3r 5 L4 (Kp 2 Ka )g
(9.24)
(9.25)
s4r 5 s5r 1 gL4 (Kp 2 Ka )
(9.26)
s5r 5 gLKp 1 gL3 (Kp 2 Ka )
(9.27)
L3 5
s2r
LKa
5
g(Kp 2 Ka )
(Kp 2 Ka )
P 5 12s2r L 1 12s2r L3
z 5 L3 1
(9.28)
(9.29)
L(2Ka 1 Kp )
LKa
L
L
5
1 5
3
Kp 2 Ka
3
3(Kp 2 Ka )
(9.30)
and Eq. (9.16) transforms to
L44 1 A 1r L34 2 A 2r L24 2 A 3r L4 2 A 4r 5 0
(9.31)
where
A 1r 5
s5r
g(Kp 2 Ka )
(9.32)
A 2r 5
8P
g(Kp 2 Ka )
(9.33)
A 3r 5
A 4r 5
6P32zg(Kp 2 Ka ) 1 s5r 4
g2 (Kp 2 Ka ) 2
P(6zs5r 1 4P)
g2 (Kp 2 Ka ) 2
(9.34)
(9.35)
451
9.5 Special Cases for Cantilever Walls Penetrating a Sandy Soil
P
L
Sand
␥
␾
c = 0
D
L5
␴3 = ␥D (Kp – Ka)
␴4 = ␥D (Kp – Ka)
Figure 9.11 Free cantilever sheet
piling penetrating a layer of sand
Free Cantilever Sheet Piling
Figure 9.11 shows a free cantilever sheet-pile wall penetrating a sandy soil and subjected
to a line load of P per unit length of the wall. For this case,
D4 2 c
2
8P
12PL
2P
d D2 2 c
dD 2 c
d 50
g(Kp 2 Ka )
g(Kp 2 Ka )
g(Kp 2 Ka )
L5 5
g(Kp 2 Ka )D2 2 2P
2D(Kp 2 Ka )g
Mmax 5 P(L 1 z r ) 2
gz93 (Kp 2 Ka )
6
(9.36)
(9.37)
(9.38)
and
zr 5
2P
Å gr (Kp 2 Ka )
(9.39)
Example 9.2
Redo parts a and b of Example 9.1, assuming the absence of the water table. Use ␥ ⫽
15.9 kN/m3 and ␾⬘ ⫽ 32°. Note: L ⫽ 5 m.
452 Chapter 9: Sheet Pile Walls
Solution
Part a
Quantity
required
Eq.
no.
Equation and calculation
Ka
—
tan2 a45 2
fr
32
b 5 tan2 a45 2 b 5 0.307
2
2
Kp
—
tan2 a45 1
fr
32
b 5 tan2 a45 1 b 5 3.25
2
2
␴2⬘
9.24
␥LKa ⫽ (15.9)(5)(0.307) ⫽ 24.41 kN/m2
L3
9.28
(5) (0.307)
LKa
5
5 0.521 m
Kp 2 Ka
3.25 2 0.307
␴5⬘
9.27
␥LKp ⫹ ␥L3(Kp ⫺ Ka) ⫽ (15.9)(5)(3.25) ⫹ (15.9)(0.521)(3.25 ⫺ 0.307)
⫽ 282.76 kN/m2
P
9.29
1
2
z
9.30
A1⬘
9.32
A2⬘
9.33
A3⬘
9.34
s2r L 1 12 s2r L3 5 12 s2r (L 1 L3 ) 5 ( 12 ) (24.41) (5 1 0.521) 5 67.38 kN>m
L(2Ka 2 Kp )
9.35
L4
9.31
53 (2) (0.307) 1 3.254
5 2.188 m
3(Kp 2 Ka )
3(3.25 2 0.307)
s5r
282.76
5
5 6.04
g(Kp 2 Ka )
(15.9) (3.25 2 0.307)
(8) (67.38)
8P
5
5 11.52
g(Kp 2 Ka )
(15.9) (3.25 2 0.307)
6P32zg(Kp 2 Ka ) 1 sr54
g2 (Kp 2 Ka ) 2
5
A4⬘
5
(6) (67.38) 3(2) (2.188) (15.9) (3.25 2 0.307) 1 282.764
(15.9) 2 (3.25 2 0.307) 2
P(6zs5r 1 4P)
2
g (Kp 2 Ka )
2
5
5 90.01
(67.38) 3(6) (2.188) (282.76) 1 (4) (67.38)4
(15.9) 2 (3.25 2 0.307) 2
5 122.52
L44 ⫹ A1⬘L43 ⫺ A2⬘L42 ⫺ A3⬘L4 ⫺ A4⬘ ⫽ 0
L44 ⫹ 6.04L43 ⫺ 11.52L24 ⫺ 90.01L4 ⫺ 122.52 ⫽ 0; L4 ⬇ 4.1 m
Dtheory ⫽ L3 ⫹ L4 ⫽ 0.521 ⫹ 4.1 ⬇ 4.7 m
Part b
Total length, L ⫹ 1.3(Dtheory) ⫽ 5 ⫹ 1.3(4.7) ⫽ 11.11 m
9.6
■
Cantilever Sheet Piling Penetrating Clay
At times, cantilever sheet piles must be driven into a clay layer possessing an undrained
cohesion c(f 5 0). The net pressure diagram will be somewhat different from that
shown in Figure 9.8a. Figure 9.12 shows a cantilever sheet-pile wall driven into clay
with a backfill of granular soil above the level of the dredge line. The water table is at
9.6 Cantilever Sheet Piling Penetrating Clay
453
A
Sand
␥
␾
c=0
L1
Water
table
␴1
C
Sand
␥sat
␾
c=0
z
L2
P1
z1
Dredge line
F
␴2
E
D
␴6
L3
Clay
␥sat
␾=0
c
G
D
z
L4
I
B
␴7
H
Figure 9.12 Cantilever sheet pile penetrating clay
a depth L1 below the top of the wall. As before, Eqs. (9.1) and (9.2) give the intensity
of the net pressures s1r and s2r , and the diagram for pressure distribution above the level
of the dredge line can be drawn. The diagram for net pressure distribution below the
dredge line can now be determined as follows.
At any depth greater than L1 1 L2 , for f 5 0, the Rankine active earth-pressure
coefficient Ka 5 1. Similarly, for f 5 0, the Rankine passive earth-pressure coefficient
Kp 5 1. Thus, above the point of rotation (point O in Figure 9.7a), the active pressure,
from right to left is
sa 5 3gL1 1 grL2 1 gsat (z 2 L1 2 L2 )4 2 2c
(9.40)
Similarly, the passive pressure from left to right may be expressed as
sp 5 gsat (z 2 L1 2 L2 ) 1 2c
(9.41)
Thus, the net pressure is
s6 5 sp 2 sa 5 3gsat (z 2 L1 2 L2 ) 1 2c4
2 3gL1 1 grL2 1 gsat (z 2 L1 2 L2 ) 4 1 2c
5 4c 2 (gL1 1 grL2 )
(9.42)
At the bottom of the sheet pile, the passive pressure from right to left is
sp 5 (gL1 1 grL2 1 gsatD) 1 2c
(9.43)
454 Chapter 9: Sheet Pile Walls
Similarly, the active pressure from left to right is
sa 5 gsatD 2 2c
(9.44)
s7 5 sp 2 sa 5 4c 1 (gL1 1 grL2 )
(9.45)
Hence, the net pressure is
For equilibrium analysis, SFH 5 0; that is, the area of the pressure diagram ACDE
minus the area of EFIB plus the area of GIH 5 0, or
P1 2 34c 2 (gL1 1 grL2 ) 4D 1 12L4 34c 2 (gL1 1 grL2 ) 1 4c 1 (gL1 1 grL2 )4 5 0
where P1 5 area of the pressure diagram ACDE.
Simplifying the preceding equation produces
L4 5
D34c 2 (gL1 1 grL2 ) 4 2 P1
4c
(9.46)
Now, taking the moment about point B (SMB 5 0) yields
P1 (D 1 z1 ) 2 34c 2 (gL1 1 grL2 )4
L4
D2
1
1 L4 (8c) ¢ ≤ 5 0
2
2
3
(9.47)
where z1 5 distance of the center of pressure of the pressure diagram ACDE,
measured from the level of the dredge line.
Combining Eqs. (9.46) and (9.47) yields
D2 34c 2 (gL1 1 grL2 ) 4 2 2DP1 2
P1 (P1 1 12cz1 )
50
(gL1 1 grL2 ) 1 2c
(9.48)
Equation (9.48) may be solved to obtain D, the theoretical depth of penetration of the clay
layer by the sheet pile.
Step-by-Step Procedure for Obtaining the Pressure Diagram
Step 1. Calculate Ka 5 tan2 (45 2 fr>2) for the granular soil (backfill).
Step 2. Obtain s1r and s2r . [See Eqs. (9.1) and (9.2).]
Step 3. Calculate P1 and z1 .
Step 4. Use Eq. (9.48) to obtain the theoretical value of D.
Step 5. Using Eq. (9.46), calculate L4 .
Step 6. Calculate s6 and s7 . [See Eqs. (9.42) and (9.45).]
Step 7. Draw the pressure distribution diagram as shown in Figure 9.12.
Step 8. The actual depth of penetration is
Dactual 5 1.4 to 1.6(Dtheoretical )
9.6 Cantilever Sheet Piling Penetrating Clay
455
Maximum Bending Moment
According to Figure 9.12, the maximum moment (zero shear) will be between
L1 1 L2 , z , L1 1 L2 1 L3 . Using a new coordinate system zr (with zr 5 0 at the
dredge line) for zero shear gives
P1 2 s6zr 5 0
or
zr 5
P1
s6
(9.49)
The magnitude of the maximum moment may now be obtained:
Mmax 5 P1 (zr 1 z1 ) 2
s6zr2
2
(9.50)
Knowing the maximum bending moment, we determine the section modulus of the sheet
pile section from Eq. (9.23).
Example 9.3
In Figure 9.13, for the sheet pile wall, determine
a. The theoretical and actual depth of penetration. Use Dactual 5 1.5Dtheory .
b. The minimum size of sheet pile section necessary. Use sall 5 172.5 MN>m2.
A
L1 = 2 m
Sand
␥ = 15.9 kN/m3
c = 0
␾ = 32°
L2 = 3 m
Sand
␥sat = 19.33 kN/m3
c = 0
␾ = 32°
D
Clay
c = 47 kN/m2
␾=0
Water table
E
B
Figure 9.13 Cantilever sheet pile penetrating into saturated clay
456 Chapter 9: Sheet Pile Walls
Solution
We will follow the step-by-step procedure given in Section 9.6:
Step 1.
Ka 5 tan2 ¢ 45 2
fr
32
≤ 5 tan2 ¢45 2 ≤ 5 0.307
2
2
Step 2.
s1r 5 gL1Ka 5 (15.9) (2) (0.307) 5 9.763 kN>m2
s2r 5 (gL1 1 grL2 )Ka 5 3(15.9) (2) 1 (19.33 2 9.81)340.307
5 18.53 kN>m2
Step 3.
From the net pressure distribution diagram given in Figure 9.12, we have
1
1
P1 5 s1r L1 1 s1r L2 1 (s2r 2 s1r )L2
2
2
5 9.763 1 29.289 1 13.151 5 52.2 kN>m
and
1
2
3
3
B9.763¢3 1 ≤ 1 29.289¢ ≤ 1 13.151¢ ≤ R
52.2
3
2
3
5 1.78 m
z1 5
Step 4.
From Eq. (9.48),
D2 34c 2 (gL1 1 grL2 )4 2 2DP1 2
P1 (P1 1 12cz1 )
50
(gL1 1 grL2 ) 1 2c
Substituting proper values yields
D2 5 (4) (47) 2 3(2) (15.9) 1 (19.33 2 9.81)346 2 2D(52.2)
52.2352.2 1 (12) (47) (1.78)4
2
50
3(15.9) (2) 1 (19.33 2 9.81)34 1 (2) (47)
or
127.64D2 2 104.4D 2 357.15 5 0
Step 5.
Solving the preceding equation, we obtain D 5 2.13 m.
From Eq. (9.46),
L4 5
D34c 2 (gL1 1 grL2 ) 4 2 P1
4c
and
4c 2 (gL1 1 grL2 ) 5 (4) (47) 2 3(15.9) (2) 1 (19.33 2 9.81)34
5 127.64 kN>m2
457
9.7 Special Cases for Cantilever Walls Penetrating Clay
So,
L4 5
2.13(127.64) 2 52.2
5 1.17 m
(4) (47)
Step 6.
s6 5 4c 2 (gL1 1 grL2 ) 5 127.64 kN>m2
s7 5 4c 1 (gL1 1 grL2 ) 5 248.36 kN>m2
Step 7.
Step 8.
The net pressure distribution diagram can now be drawn, as shown in
Figure 9.12.
Dactual < 1.5Dtheoretical 5 1.5(2.13) < 3.2 m
Maximum-Moment Calculation
From Eq. (9.49),
zr 5
P1
52.2
5
< 0.41 m
s6
127.64
Again, from Eq. (9.50),
Mmax 5 P1 (zr 1 z1 ) 2
s6zr2
2
So
127.64(0.41) 2
2
5 114.32 2 10.73 5 103.59 kN-m>m
Mmax 5 52.2(0.41 1 1.78) 2
The minimum required section modulus (assuming that sall 5 172.5 MN>m2) is
S5
9.7
103.59 kN-m>m
5 0.6 3 1023 m3 , m of the wall
172.5 3 103 kN>m2
■
Special Cases for Cantilever Walls Penetrating Clay
Sheet Pile Wall in the Absence of Water Table
As in Section 9.5, relationships for special cases for cantilever walls penetrating clay may
also be derived. Referring to Figure 9.14, we can write
s2r 5 gLKa
(9.51)
s6 5 4c 2 gL
(9.52)
s7 5 4c 1 gL
(9.53)
P1 5
1
2 Ls2r
1
2
2 gL Ka
(9.54)
L4 5
D(4c 2 gL) 2 12gL2Ka
4c
(9.55)
5
and
458 Chapter 9: Sheet Pile Walls
Sand
␥
␾
L
P1
z1
␴2
␴6
Clay
L3
␥sat
␾ = 0
c
D
L4
␴7
Figure 9.14 Sheet-pile wall penetrating clay
The theoretical depth of penetration, D, can be calculated [in a manner similar to the calculation of Eq. (9.48)] as
D2 (4c 2 gL) 2 2DP1 2
where z1 5
P1 (P1 1 12cz1 )
50
gL 1 2c
L
.
3
(9.56)
(9.57)
The magnitude of the maximum moment in the wall is
Mmax 5 P1 (zr 1 z1 ) 2
where zr 5
s6zr2
2
1
2
P1
2 gL Ka
.
5
s6
4c 2 gL
(9.58)
(9.59)
Free Cantilever Sheet-Pile Wall Penetrating Clay
Figure 9.15 shows a free cantilever sheet-pile wall penetrating a clay layer. The wall is
being subjected to a line load of P per unit length. For this case,
␴6 ⫽ ␴7 ⫽ 4c
(9.60)
The depth of penetration, D, may be obtained from the relation
4D2c 2 2PD 2
P(P 1 12cL)
50
2c
(9.61)
9.7 Special Cases for Cantilever Walls Penetrating Clay
459
P
L
␴6
L3
Clay
D
L4
␥sat
␾ = 0
c
␴7
Figure 9.15 Free cantilever sheet piling penetrating clay
Also, note that, for a construction of the pressure diagram,
4cD 2 P
4c
The maximum moment in the wall is
L4 5
Mmax 5 P(L 1 z r ) 2
where
zr 5
(9.62)
4cz r2
2
P
4c
(9.63)
(9.64)
Example 9.4
Refer to the free cantilever sheet-pile wall shown in Figure 9.15, for which P ⫽ 32
kN/m, L ⫽ 3.5 m, and c ⫽ 12 kN/m2. Calculate the theoretical depth of penetration.
Solution
From Eq. (9.61),
P(P 1 12cL)
50
2c
32332 1 (12) (12) (3.5)4
(4) (D2 ) (12) 2 (2) (32) (D) 2
50
(2) (12)
48D2 ⫺ 64D ⫺ 714.7 ⫽ 0
4D2c 2 2PD 2
Hence D ⬇ 4.6 m.
■
460 Chapter 9: Sheet Pile Walls
9.8
Anchored Sheet-Pile Walls
When the height of the backfill material behind a cantilever sheet-pile wall exceeds
about 6 m, tying the wall near the top to anchor plates, anchor walls, or anchor piles
becomes more economical. This type of construction is referred to as anchored sheetpile wall or an anchored bulkhead. Anchors minimize the depth of penetration
required by the sheet piles and also reduce the cross-sectional area and weight of the
sheet piles needed for construction. However, the tie rods and anchors must be carefully designed.
The two basic methods of designing anchored sheet-pile walls are (a) the free earth
support method and (b) the fixed earth support method. Figure 9.16 shows the assumed
nature of deflection of the sheet piles for the two methods.
Anchor tie rod
Water
table
Moment
Mmax
Dredge line
D
Sheet pile
simply supported
(a)
Anchor tie rod
Moment
Water
table
Mmax
Deflection
Dredge line
Point of inflection
D
Sheet pile fixed
at lower end
(b)
Figure 9.16 Nature of variation of deflection and moment for anchored sheet piles:
(a) free earth support method; (b) fixed earth support method
9.9 Free Earth Support Method for Penetration of Sandy Soil
461
The free earth support method involves a minimum penetration depth. Below the
dredge line, no pivot point exists for the static system. The nature of the variation of the
bending moment with depth for both methods is also shown in Figure 9.16. Note that
Dfree earth , Dfixed earth
9.9
Free Earth Support Method for Penetration
of Sandy Soil
Figure 9.17 shows an anchor sheet-pile wall with a granular soil backfill; the wall has been
driven into a granular soil. The tie rod connecting the sheet pile and the anchor is located
at a depth l1 below the top of the sheet-pile wall.
The diagram of the net pressure distribution above the dredge line is similar to that
shown in Figure 9.8. At depth z 5 L1 , s1r 5 gL1Ka , and at z 5 L1 1 L2 , s2r 5
(gL1 1 grL2 )Ka . Below the dredge line, the net pressure will be zero at
z 5 L1 1 L2 1 L3 . The relation for L3 is given by Eq. (9.6), or
L3 5
s2r
gr(Kp 2 Ka )
A
Anchor tie rod
L1 O
Water
table
␴1
Water
table
C
l1
F
Sand
␥, ␾
l2
z
L2
P
z
␴2
Dredge line
L3
D
1
E
D
␥(Kp – Ka)
L4
F
␴8
Sand
␥sat, ␾
B
Figure 9.17 Anchored sheet-pile wall penetrating sand
Sand
␥sat, ␾
462 Chapter 9: Sheet Pile Walls
At z 5 L1 1 L2 1 L3 1 L4 , the net pressure is given by
s8r 5 gr(Kp 2 Ka )L4
(9.65)
Note that the slope of the line DEF is 1 vertical to gr(Kp 2 Ka ) horizontal.
For equilibrium of the sheet pile, S horizontal forces 5 0, and S moment about
Or 5 0. (Note: Point Or is located at the level of the tie rod.)
Summing the forces in the horizontal direction (per unit length of the wall) gives
Area of the pressure diagram ACDE 2 area of EBF 2 F 5 0
where F 5 tension in the tie rod> unit length of the wall, or
P 2 12 s8r L4 2 F 5 0
or
F 5 P 2 12 3 gr(Kp 2 Ka )4L24
(9.66)
where P 5 area of the pressure diagram ACDE. Now, taking the moment about point Or
gives
2P3(L1 1 L2 1 L3 ) 2 (z 1 l1 )4 1 12 3gr(Kp 2 Ka )4L24 (l2 1 L2 1 L3 1 23L4 ) 5 0
or
L34 1 1.5L24 (l2 1 L2 1 L3 ) 2
3P3(L1 1 L2 1 L3 ) 2 (z 1 l1 )4
gr (Kp 2 Ka )
50
(9.67)
Equation (9.67) may be solved by trial and error to determine the theoretical depth, L4 :
Dtheoretical 5 L3 1 L4
The theoretical depth is increased by about 30 to 40% for actual construction, or
Dactual 5 1.3 to 1.4Dtheoretical
(9.68)
The step-by-step procedure in Section 9.4 indicated that a factor of safety can be
applied to Kp at the beginning [i.e., Kp(design) 5 Kp>FS]. If that is done, there is no need to
increase the theoretical depth by 30 to 40%. This approach is often more conservative.
9.9 Free Earth Support Method for Penetration of Sandy Soil
463
The maximum theoretical moment to which the sheet pile will be subjected occurs
at a depth between z 5 L1 and z 5 L1 1 L2 . The depth z for zero shear and hence maximum moment may be evaluated from
1
2 s1r L1
2 F 1 s1r (z 2 L1 ) 1 12Kagr(z 2 L1 ) 2 5 0
(9.69)
Once the value of z is determined, the magnitude of the maximum moment is easily obtained.
Example 9.5
Let L1 ⫽ 3.05 m, L2 ⫽ 6.1 m, l1 ⫽ 1.53 m, l2 ⫽ 1.52 m, c⬘ ⫽ 0, ␾⬘ ⫽ 30°, ␥ ⫽ 16
kN/m3, ␥sat ⫽ 19.5 kN/m3, and E ⫽ 207 ⫻ 103 MN/m2 in Figure 9.17.
a. Determine the theoretical and actual depths of penetration. (Note: Dactual ⫽
1.3Dtheory.)
b. Find the anchor force per unit length of the wall.
c. Determine the maximum moment, Mmax.
Solution
Part a
We use the following table.
Quantity
required
Eq.
no.
Ka
—
tan2 a45 2
fr
30
1
b 5 tan2 a45 2 b 5
2
2
3
KP
—
tan2 a45 1
fr
30
b 5 tan2 a45 1 b 5 3
2
2
Kp ⫺ Ka
␥⬘
—
—
3 ⫺ 0.333 ⫽ 2.667
␥sat ⫺ ␥w ⫽ 19.5 ⫺ 9.81 ⫽ 9.69 kN/m3
␴1⬘
9.1
gL1Ka 5 (16) (3.05) ( 13 ) 5 16.27 kN>m2
␴2⬘
9.2
L3
9.6
s2r
35.97
5
5 1.39 m
gr (Kp 2 Ka )
(9.69) (2.667)
P
—
1
2 s1r L1
Equation and calculation
(gL1 1 grL2 )Ka 5 3 (16) (3.05) 1 (9.69) (6.1)4 13 5 35.97 kN>m2
1 s2r L2 1 12 (s2r 2 s1r )L2 1 12s2r L3 5 ( 12 ) (16.27) (3.05)
1 (16.27) (6.1) 1 ( 12 ) (35.97 2 16.27) (6.1) 1 ( 12 ) (35.97) (1.39)
5 24.81 1 99.25 1 60.01 1 25.0 5 209.07 kN>m
z
—
3.05
6.1
b 1 (99.25) a1.39 1
b
3
2
1
¥
6.1
2 3 1.39
209.07
1 (60.01) a1.39 1
b 1 (25.0) a
b
3
3
5 4.21 m
SME
5 ≥
P
(24.81) a1.39 1 6.1 1
464 Chapter 9: Sheet Pile Walls
L4
L34 1 1.5L24 (l2 1 L2 1 L3 ) 2
9.67
3P3 (L1 1 L2 1 L3 ) 2 (z 1 l1 )4
gr(Kp 2 Ka )
50
L43 ⫹ 1.5L42(1.52 ⫹ 6.1 ⫹ 1.39)
2
(3) (209.07) 3 (3.05 1 6.1 1 1.39) 2 (4.21 1 1.53)4
(9.69) (2.667)
50
L4 5 2.7 m
Dtheory
—
L3 ⫹ L4 ⫽ 1.39 ⫹ 2.7 ⫽ 4.09 ⬇ 4.1 m
Dactual
—
1.3Dtheory ⫽ (1.3)(4.1) ⫽ 5.33 m
Part b
The anchor force per unit length of the wall is
F 5 P 2 12gr(Kp 2 Ka )L24
5 209.07 2 A 12 B (9.69) (2.667) (2.7) 2 5 114.87 kN>m < 115 kN>m
Part c
From Eq. (9.69), for zero shear,
1
2
s1r L1 2 F 1 s1r (z 2 L1 ) 1 12 Kagr (z 2 L1 ) 2 5 0
Let z ⫺ L1 ⫽ x, so that
1
2
s1r l1 2 F 1 s1r x 1 12 Kagrx2 5 0
or
A 12 B (16.27) (3.05) 2 115 1 (16.27) (x) 1 A 12 BA 13 B (9.69)x2 5 0
giving
x2 ⫹ 10.07x ⫺ 55.84 ⫽ 0
Now, x ⫽ 4 m and z ⫽ x ⫹ L1 ⫽ 4 ⫹ 3.05 ⫽ 7.05 m. Taking the moment about the point
of zero shear, we obtain
1
1
3.05
x2
x
b 1 F(x 1 1.52) 2 s1r 2 Kagrx2 a b
Mmax 5 2 s1r L1 ax 1
2
3
2
2
3
or
1
3.05
42
Mmax 5 2 a b (16.27) (3.05) a4 1
b 1 (115) (4 1 1.52) 2 (16.27) a b
2
3
2
1 1
4
2 a b a b (9.69) (4) 2 a b 5 344.9 kN ? m>m
2 3
3
■
9.10 Design Charts for Free Earth Support Method (Penetration into Sandy Soil)
9.10
465
Design Charts for Free Earth Support Method
(Penetration into Sandy Soil)
Using the free earth support method, Hagerty and Nofal (1992) provided simplified design
charts for quick estimation of the depth of penetration, D, anchor force, F, and maximum
moment, Mmax, for anchored sheet-pile walls penetrating into sandy soil, as shown in
Figure 9.17. They made the following assumptions for their analysis.
a. The soil friction angle, fr, above and below the dredge line is the same.
b. The angle of friction between the sheet-pile wall and the soil is fr>2.
c. The passive earth pressure below the dredge line has a logarithmic spiral failure
surface.
d. For active earth-pressure calculation, Coulomb’s theory is valid.
The magnitudes of D, F, and Mmax may be calculated from the following relationships:
D
5 (GD) (CDL1 )
L1 1 L2
(9.70)
F
5 (GF) (CFL1 )
ga (L1 1 L2 ) 2
(9.71)
Mmax
ga (L1 1 L2 ) 3
5 (GM) (CML1 )
(9.72)
where
ga 5 average unit weight of soil
5
gL21 1 (gsat 2 gw )L22 1 2gL1L2
(L1 1 L2 ) 2
GD 5 generalized nondimensional embedment
5
D
L1 1 L2
(for L1 5 0 and L2 5 L1 1 L2 )
GF 5 generalized nondimensional anchor force
5
F
ga (L1 1 L2 ) 2
(for L1 5 0 and L2 5 L1 1 L2 )
(9.73)
466 Chapter 9: Sheet Pile Walls
GM 5 generalized nondimensional moment
5
Mmax
ga (L1 1 L2 ) 3
(for L1 5 0 and L2 5 L1 1 L2 )
CDL1, CFL1, CML1 5 correction factors for L1 2 0
The variations of GD, GF, GM, CDL1, CFL1, and CML1 are shown in Figures 9.18,
9.19, 9.20, 9.21, 9.22, and 9.23, respectively.
0.5
0.4
GD
24 ␾
26°
0.3
28°
30°
32°
0.2
34°
36°
38°
0.1
0.0
0.1
0.2
0.3
l1/(L1 L2)
0.4
0.5
Figure 9.18 Variation of GD with
l1> (L1 1 L2 ) and fr (Hagerty, D.
J., and Nofal, M. M. (1992).
“Design Aids: Anchored Bulkheads
in Sand,” Canadian Geotechnical
Journal, Vol. 29, No. 5,
pp. 789–795. © 2008 NRC Canada
or its licensors. Reproduced with
permission.)
0.16
0.14
24 ␾
0.12
26°
GF
28°
0.10
30°
32°
34°
36°
38°
0.08
0.06
0.04
0.0
0.1
0.2
0.3
l1/(L1 L2)
0.4
0.5
Figure 9.19 Variation of GF
with l1> (L1 1 L2 ) and fr (After
Hagerty and Nofal, 1992)
(Hagerty, D. J., and Nofal, M. M.
(1992). "Design Aids: Anchored
Bulkheads in Sand," Canadian
Geotechnical Journal, Vol. 29,
No. 5, pp. 789-795. © 2008 NRC
Canada or its licensors.
Reproduced with permission.
0.05
0.04
GM
0.03
24 ␾
26°
28°
30°
0.02
0.01
32°
34°
36°
38°
0.00
0.0
0.1
0.2
0.3
l1/(L1 L2)
0.4
0.5
Figure 9.20 Variation of GM with l1> (L1 1 L2 )
and fr (Hagerty, D. J., and Nofal, M. M. (1992).
“Design Aids: Anchored Bulkheads in Sand,”
Canadian Geotechnical Journal, Vol. 29, No. 5,
pp. 789–795. © 2008 NRC Canada or its licensors. Reproduced with permission.)
1.18
1.16
L1
0.4
L1 L2
1.14
CDL1
1.12
0.3
1.10
1.08
0.2
1.06
1.04
0.1
0.0
0.1
0.2
0.3
l1/(L1 L2)
0.4
0.5
Figure 9.21 Variation of CDL1 with L1> (L1 1 L2 ) and
l1> (L1 1 L2 ) (Hagerty, D. J., and Nofal, M. M. (1992).
“Design Aids: Anchored Bulkheads in Sand,” Canadian
Geotechnical Journal, Vol. 29, No. 5, pp. 789–795.
© 2008 NRC Canada or its licensors. Reproduced
with permission.)
1.08
1.07
L1
0.4
L1 L2
0.3
CFL1
1.06
0.2
1.05
1.04
0.1
1.03
0.0
0.1
0.2
0.3
l1/(L1 L2)
0.4
0.5
Figure 9.22 Variation of CFL1 with
L1> (L1 1 L2 ) and l1> (L1 1 L2 )
(Hagerty, D. J., and Nofal, M. M.
(1992). “Design Aids: Anchored
Bulkheads in Sand,” Canadian
Geotechnical Journal, Vol. 29,
No. 5, pp. 789–795. © 2008 NRC
Canada or its licensors. Reproduced
with permission.)
467
468 Chapter 9: Sheet Pile Walls
1.06
1.04
CML1
1.02
1.00
L1
0.4
L1 L2
0.1
0.98
0.3
0.1
0.2
0.96
0.94
0.0
0.1
0.2
0.3
l1/(L1 L2)
0.4
0.5
Figure 9.23 Variation of CML1 with L1> (L1 1 L2 ) and l1> (L1 1 L2 )
(Hagerty, D. J., and Nofal, M. M. (1992). “Design Aids: Anchored Bulkheads in
Sand,” Canadian Geotechnical Journal, Vol. 29, No. 5, pp. 789–795. © 2008
NRC Canada or its licensors. Reproduced with permission.)
Example 9.6
Refer to Figure 9.17. Given: L1 ⫽ 2 m, L2 ⫽ 3 m, l1 ⫽ l2 ⫽ 1 m, c ⫽ 0, ␾⬘ ⫽ 32° ␥ ⫽
15.9 kN/m3, and ␥sat ⫽ 19.33 kN/m3. Determine:
a. Theoretical and actual depth of penetration. Note: Dactual ⫽ 1.4Dtheory.
b. Anchor force per unit length of wall.
c. Maximum moment, Mmax.
Use the charts presented in Section 9.10.
Solution
Part a
From Eq. (9.70),
D
5 (GD) (CDL1 )
L1 1 L2
l1
1
5
5 0.2
L1 1 L2
213
From Figure 9.18 for l1/(L1 ⫹ L2) ⫽ 0.2 and ␾⬘ ⫽ 32°, GD ⫽ 0.22. From Figure 9.21, for
L1
2
5
5 0.4
L1 1 L2
213
and
l1
5 0.2
L1 1 L2
9.11 Moment Reduction for Anchored Sheet-Pile Walls
469
CDL1 ⬇ 1.172. So
Dtheory ⫽ (L1 ⫹ L2)(GD)(CDL1) ⫽ (5)(0.22)(1.172) ⬇ 1.3
Dactual ⬇ (1.4)(1.3) ⫽ 1.82 ⬇ 2 m
Part b
From Figure 9.19 for l1/(L1 ⫹ L2) ⫽ 0.2 and ␾⬘ ⫽ 32°, GF ⬇ 0.074. Also, from
Figure 9.22, for
L1
2
5
5 0.4,
L1 1 L2
213
l1
5 0.2,
L1 1 L2
and fr 5 32°
CFL1 ⫽ 1.073. From Eq. (9.73),
ga 5
5
gL21 1 g9L22 1 2gL1L2
(L1 1 L2 ) 2
(15.9) (2) 2 1 (19.33 2 9.81) (3) 2 1 (2) (15.9) (2) (3)
(2 1 3) 2
5 13.6 kN>m3
Using Eq. (9.71) yields
F ⫽ ␥a(L1 ⫹ L2)2(GF)(CFL1) ⫽ (13.6)(5)2(0.074)(1.073) ⬇ 27 kN/m
Part c
From Figure 9.20, for l1/(L1 ⫹ L2) ⫽ 0.2 and ␾ ⫽ 32°, GM ⫽ 0.021. Also, from
Figure 9.23, for
L1
2
5
5 0.4,
L1 1 L2
213
l1
5 0.2,
L1 1 L2
and fr 5 32°
CML1 ⫽ 1.036. Hence from Eq. (9.72),
Mmax ⫽ ␥a(L1 ⫹ L2)3(GM)(CML1) ⫽ (13.6)(5)3(0.021)(1.036) ⫽ 36.99 kN ⭈ m/m
9.11
■
Moment Reduction for Anchored Sheet-Pile Walls
Sheet piles are flexible, and hence sheet-pile walls yield (i.e., become displaced
laterally), which redistributes the lateral earth pressure. This change tends to reduce
the maximum bending moment, Mmax , as calculated by the procedure outlined in
470 Chapter 9: Sheet Pile Walls
1.0
Loose
sand
␣H
0.8
H L1 L2 Dactual
Safe
section
Md
Mmax
0.6
Dense sand
and gravel
0.4
Unsafe section
0.2
Flexible
piles
Stiff
piles
0
4.0
3.5
3.0
Log ρ
2.5
2.0
Figure 9.24 Plot of log r against Md>Mmax for sheet-pile walls penetrating sand
(From Rowe, P. W. (1952). “Anchored Sheet Pile Walls,” Proceedings, Institute of Civil Engineers,
Vol. 1, Part 1, pp. 27–70. )
Section 9.9. For that reason, Rowe (1952, 1957) suggested a procedure for reducing the
maximum design moment on the sheet pile walls obtained from the free earth support
method. This section discusses the procedure of moment reduction for sheet piles penetrating into sand.
In Figure 9.24, which is valid for the case of a sheet pile penetrating sand, the following notation is used:
1. Hr 5 total height of pile driven (i.e., L1 1 L2 1 Dactual)
4
2.
Hr
Relative flexibility of pile 5 r 5 10.91 3 10 ¢
≤
EI
27
(9.74)
where
Hr is in meters
E 5 modulus of elasticity of the pile material (MN> m2 )
I 5 moment of inertia of the pile section per meter of the wall (m4>m of wall)
3. Md 5 design moment
4. Mmax 5 maximum theoretical moment
9.11 Moment Reduction for Anchored Sheet-Pile Walls
471
The procedure for the use of the moment reduction diagram (see Figure 9.24) is as
follows:
Step 1. Choose a sheet pile section (e.g., from among those given in Table 9.1).
Step 2. Find the modulus S of the selected section (Step 1) per unit length of
the wall.
Step 3. Determine the moment of inertia of the section (Step 1) per unit length of
the wall.
Step 4. Obtain Hr and calculate r [see Eq. (9.74)].
Step 5. Find log r.
Step 6. Find the moment capacity of the pile section chosen in Step 1 as
Md 5 sallS.
Step 7. Determine Md>Mmax . Note that Mmax is the maximum theoretical moment
determined before.
Step 8. Plot log r (Step 5) and Md>Mmax in Figure 9.24.
Step 9. Repeat Steps 1 through 8 for several sections. The points that fall above the
curve (in loose sand or dense sand, as the case may be) are safe sections.
The points that fall below the curve are unsafe sections. The cheapest section may now be chosen from those points which fall above the proper
curve. Note that the section chosen will have an Md , Mmax .
Example 9.7
Refer to Example 9.5. Use Rowe’s moment reduction diagram (Figure 9.24) to find an appropriate sheet pile section. For the sheet pile, use E 5 207 3 103 MN>m2 and
sall 5 172,500 kN>m2.
Solution
H r 5 L1 1 L2 1 Dactual 5 3.05 1 6.1 1 5.33 5 14.48 m
Mmax 5 344.9 kN ? m>m. Now the following table can be prepared.
Section
PZ-22
PZ-27
4
l (m /m)
r 5 10.91 3
Hr4
1027 a
b
H⬘(m)
El
115.2 ⫻ 10⫺6 14.48
251.5 ⫻ 10⫺6 14.48
20.11 ⫻ 10⫺4
9.21 ⫻ 10⫺4
␴all
Md ⫽ S␴
log ␳
3
S(m /m)
⫺2.7
97 ⫻ 10⫺5
⫺3.04 162.3 ⫻ 10⫺5
(kN ⭈ m /m)
Md
Mmax
167.33
284.84
0.485
0.826
Figure 9.25 gives a plot of Md/Mmax versus ␳. It can be seen that PZ-27 will be
sufficient.
472 Chapter 9: Sheet Pile Walls
1.0
PZ-27
0.8
Md
Mmax
0.6
Loose sand
PZ-22
0.4
0.2
0
–4.0
–3.5
–3.0
Log ␳
–2.5
–2.0
Figure 9.25 Plot of Md/Mmax versus log ␳
9.12
■
Computational Pressure Diagram Method
for Penetration into Sandy Soil
The computational pressure diagram (CPD) method for sheet pile penetrating a sandy soil is
a simplified method of design and an alternative to the free earth method described in
Sections 9.9 and 9.11 (Nataraj and Hoadley, 1984). In this method, the net pressure diagram
shown in Figure 9.17 is replaced by rectangular pressure diagrams, as in Figure 9.26. Note
that sar is the width of the net active pressure diagram above the dredge line and spr is the
width of the net passive pressure diagram below the dredge line. The magnitudes of sar and
spr may respectively be expressed as
sar 5 CKagav
rL
(9.75)
spr 5 RCKagav
r L 5 Rsar
(9.76)
and
where
gav
r 5 average effective unit weight of sand
gL1 1 grL2
<
L1 1 L2
C 5 coefficient
L(L 2 2l1 )
R 5 coefficient 5
D(2L 1 D 2 2l1 )
The range of values for C and R is given in Table 9.2.
(9.77)
(9.78)
473
9.12 Computational Pressure Diagram Method for Penetration into Sandy Soil
l1
L1
Water
table
l2
Sand; ␥, ␾
F
Anchor tie rod
Sand
L2
␥sat
␾
␴a
Sand
D
␥sat
␾
␴p
Figure 9.26 Computational pressure
diagram method (Note: L1 1 L2 5 L)
Table 9.2 Range of Values for C and R [from Eqs. (9.75) and (9.76)]
Ca
Soil type
Loose sand
Medium sand
Dense sand
R
0.8– 0.85
0.7– 0.75
0.55–0.65
0.3–0.5
0.55–0.65
0.60–0.75
a
Valid for the case in which there is no surcharge above the granular backfill (i.e., on the right
side of the wall, as shown in Figure 9.26)
The depth of penetration, D, anchor force per unit length of the wall, F, and maximum moment in the wall, Mmax , are obtained from the following relationships.
Depth of Penetration
For the depth of penetration, we have
D2 1 2DL B 1 2 ¢
l1
l1
L2
≤ R 2 ¢ ≤ B1 2 2¢ ≤ R 5 0
L
R
L
(9.79)
Anchor Force
The anchor force is
F 5 sar (L 2 RD)
(9.80)
474 Chapter 9: Sheet Pile Walls
Maximum Moment
The maximum moment is calculated from
Mmax 5 0.5 sar L2 B ¢1 2
2l1
RD 2
RD
≤ 2 ¢ ≤ ¢1 2
≤R
L
L
L
(9.81)
Note the following qualifications:
1. The magnitude of D obtained from Eq. (9.79) is about 1.25 to 1.5 times the value of
Dtheory obtained by the conventional free earth support method (see Section 9.9), so
D < Dactual
c
Eq. (9.79)
c
Eq. (9.68)
2. The magnitude of F obtained by using Eq. (9.80) is about 1.2 to 1.6 times the value
obtained by using Eq. (9.66). Thus, an additional factor of safety for the actual
design of anchors need not be used.
3. The magnitude of Mmax obtained from Eq. (9.81) is about 0.6 to 0.75 times the value
of Mmax obtained by the conventional free earth support method. Hence, the former
value of Mmax can be used as the actual design value, and Rowe’s moment reduction
need not be applied.
Example 9.8
For the anchored sheet pile wall shown in Figure 9.27, determine (a) D, (b) F, and (c)
Mmax. Use the CPD method; assume that C 5 0.68 and R 5 0.6.
Solution
Part a
gr 5 gsat 2 gw 5 19.24 2 9.81 5 9.43 kN>m3
From Eq. (9.77)
r 5
gav
gL1 1 grL2
(17.3) (3) 1 (9.43) (6)
5
5 12.05 kN>m3
L1 1 L2
316
Ka 5 tan2 ¢45 2
fr
35
≤ 5 tan2 ¢ 45 2 ≤ 5 0.271
2
2
sar 5 CKagav
r L 5 (0.68) (0.271) (12.05) (9) 5 19.99 kN>m2
spr 5 Rsar 5 (0.6) (19.99) 5 11.99 kN>m2
From Eq. (9.80)
D2 1 2DL B1 2 ¢
l1
l1
L2
≤R 2
B1 2 2¢ ≤ R 5 0
L
R
L
9.12 Computational Pressure Diagram Method for Penetration into Sandy Soil
L1 3 m
Water table
l1 1.5 m
Anchor
475
Sand
c 0
␥ 17.3 kN/m3
␾ 35°
L2 6 m
Sand
␥sat 19.24 kN/m3
c 0
␾ 35°
D
Sand
␥sat 19.24 kN/m3
c 0
␾ 35°
Figure 9.27
or
D2 1 2(D) (9) B1 2 ¢
(9) 2
1.5
1.5
≤R 2
B1 2 2¢
≤ R 5 D2 1 50D 2 1000 5 0
9
0.6
9
Hence D < 4.6 m.
Check for the assumption of R:
R5
939 2 (2) (1.5)4
L(L 2 2l1 )
5
< 0.6 —OK
D(2L 1 D 2 2l1 )
4.63(2) (9) 1 4.6 2 (2) (1.5)4
Part b
From Eq. (9.80)
F 5 sar (L 2 RD) 5 19.9939 2 (0.6) (4.6)4 5 124.74 kN , m
Part c
From Eq. (9.81)
Mmax 5 0.5sar L2 B ¢1 2
12
2l1
RD 2
RD
≤ 2 ¢ ≤ ¢1 2
≤R
L
L
L
(0.6) (4.6)
RD
512
5 0.693
L
9
So,
Mmax 5 (0.5) (19.99) (9) 2 B (0.693) 2 2
(2) (1.5) (0.693)
R 5 201.6 kN-m , m
9
■
476
Chapter 9: Sheet Pile Walls
9.13
Fixed Earth-Support Method for Penetration
into Sandy Soil
When using the fixed earth support method, we assume that the toe of the pile is restrained
from rotating, as shown in Figure 9.28a. In the fixed earth support solution, a simplified
method called the equivalent beam solution is generally used to calculate L3 and, thus, D.
The development of the equivalent beam method is generally attributed to Blum (1931).
In order to understand this method, compare the sheet pile to a loaded cantilever beam
RSTU, as shown in Figure 9.29. Note that the support at T for the beam is equivalent to the
anchor load reaction (F) on the sheet pile (Figure 9.28). It can be seen that the point S of the
beam RSTU is the inflection point of the elastic line of the beam, which is equivalent to point
I in Figure 9.28. It the beam is cut at S and a free support (reaction Ps) is provided at that
point, the bending moment diagram for portion STU of the beam will remain unchanged.
This beam STU will be equivalent to the section STU of the beam RSTU. The force P⬘ shown
in Figure 9.28a at I will be equivalent to the reaction Ps on the beam (Figure 9.29).
The following is an approximate procedure for the design of an anchored sheet-pile
wall (Cornfield, 1975). Refer to Figure 9.28.
Step 1.
Determine L5, which is a function of the soil friction angle ␾⬘ below the
dredge line, from the following:
␾ⴕ (deg)
L5
L1 1 L2
30
35
40
0.08
0.03
0
A
l1
L1
O′
␴1
Deflected
shape of
sheet pile
F
Anchor l
2
C
Water table
Sand
γ, φ′
z
Sand
γsat
φ′
L2
␴2
L5
L3
D
L5
I
J
P′
E
D
H
F
B
(a) Pressure diagram
G
(b) Moment diagram
FIGURE 9.28 Fixed earth support method for penetration of sandy soil
9.13 Fixed Earth-Support Method for Penetration into Sandy Soil
R
S
T
477
U
Beam
Ps
Moment diagram
Figure 9.29 Equivalent cantilever beam concept
Step 4.
Step 5.
Calculate the span of the equivalent beam as l2 ⫹ L2 ⫹ L5 ⫽ L⬘.
Calculate the total load of the span, W. This is the area of the pressure
diagram between O⬘ and I.
Calculate the maximum moment, Mmax, as WL⬘/8.
Calculate P⬘ by taking the moment about O⬘, or
Step 6.
1
(moment of area ACDJI about O r )
Lr
Calculate D as
Step 2.
Step 3.
Pr 5
6P r
Å (Kp 2 Ka )gr
D 5 L5 1 1.2
Step 7.
(9.82)
(9.83)
Calculate the anchor force per unit length, F, by taking the moment
about I, or
1
(moment of area ACDJI about I)
F5
Lr
Example 9.9
Consider the anchored sheet-pile structure described in Example 9.5. Using the equivalent beam method described in Section 9.13, determine
a. Maximum moment
b. Theoretical depth of penetration
c. Anchor force per unit length of the structure
Solution
Part a
Determination of L5: For ␾⬘ ⫽ 30°,
L5
5 0.08
L1 1 L2
478 Chapter 9: Sheet Pile Walls
L5
5 0.08
3.05 1 6.1
L5 ⫽ 0.73
Net Pressure Diagram: From Example 9.5, Ka 5 31 , Kp ⫽ 3, ␥ ⫽ 16 kN/m3, ␥⬘ ⫽ 9.69
kN/m3, ␴1⬘ ⫽ 16.27 kN/m2, ␴2⬘ ⫽ 35.97 kN/m2. The net active pressure at a depth L5
below the dredge line can be calculated as
␴2⬘ ⫺ ␥⬘(Kp ⫺ Ka)L5 ⫽ 35.97 ⫺ (9.69)(3 ⫺ 0.333)(0.73) ⫽ 17.1 kN/m2
The net pressure diagram from z ⫽ 0 to z ⫽ L1 ⫹ L2 ⫹ L5 is shown in Figure 9.30.
Maximum Moment:
1
1
W 5 a b (8.16 1 16.27) (1.52) 1 a b (6.1) (16.27 1 35.97)
2
2
1
1 a b (0.73) (35.97 1 17.1)
2
⫽ 197.2 kN/m
L⬘ ⫽ l2 ⫹ L2 ⫹ L5 ⫽ 1.52 ⫹ 6.1 ⫹ 0.73 ⫽ 8.35 m
Mmax 5
(197.2) (8.35)
WLr
5
5 205.8 kN ? m>m
8
8
Part b
Pr 5
1
(moment of area ACDJI about O⬘)
Lr
A
8.16 kN/m2
O′
1.53 m = l1
F
l2 = 1.52 m
16.27 kN/m2
C
6.1 m = L2
35.97 kN/m2
D
L5 = 0.73 m
P′
I
17.1 kN/m2
J
FIGURE 9.30
9.14 Field Observations for Anchor Sheet Pile Walls
479
-
£1≥
2
6.1 ≥
(16.27) (3.05) £ 3 3.05 2 1.53≥ 1 (16.27) (6.1) £1.52 1
2
3
2
1
1
2
1
P9 5
H1 £ ≥ (6.1) (35.97 2 16.27) £1.52 1 3 6.1≥ 1 £ ≥ (35.97 1 17.1) X
8.35
2
3
2
0.73
≥
3 (0.73) £1.52 1 6.1 1
2
c
Approximate
⫽ 114.48 kN/m
From Eq. (9.83)
D 5 L5 1 1.2
(6) (114.48)
6P r
5 0.73 1 1.2
5 6.92 m
Å (Kp 2 Ka )gr
Å (3 2 0.333) (9.69)
Part c
Taking the moment about I (Figure 9.30)
1
3.05
6.1
a (16.27) (3.05) a0.73 1 6.1 1
b 1 (16.27) (6.1) a0.73 1
b
1
2
3
2
F5
E
U
8.35
1
6.1
1
0.73
1 a b (6.1) (35.97 2 16.27) a0.73 1
b 1 a b (35.97 1 17.1) (0.73) a
b
2
3
2
2
c
Approximate
⫽ 88.95 kN/m
9.14
■
Field Observations for Anchor Sheet Pile Walls
In the preceding sections, large factors of safety were used for the depth of penetration, D.
In most cases, designers use smaller magnitudes of soil friction angle, ␾⬘, thereby ensuring a built-in factor of safety for the active earth pressure. This procedure is followed primarily because of the uncertainties involved in predicting the actual earth pressure to
which a sheet-pile wall in the field will be subjected. In addition, Casagrande (1973)
observed that, if the soil behind the sheet-pile wall has grain sizes that are predominantly
smaller than those of coarse sand, the active earth pressure after construction sometimes
increases to an at-rest earth-pressure condition. Such an increase causes a large increase in
the anchor force, F. The following two case histories are given by Casagrande (1973).
Bulkhead of Pier C—Long Beach Harbor, California (1949)
A typical cross section of the Pier C bulkhead of the Long Beach harbor is shown in Figure
9.31. Except for a rockfill dike constructed with 76 mm (3 in.) maximum-size quarry
wastes, the backfill of the sheet-pile wall consisted of fine sand. Figure 9.32 shows the
480 Chapter 9: Sheet Pile Walls
+5.18 m
+1.22 m
Tie rod
–76 mm dia.
Fine sand
–3.05 m
hydraulic fill
El.0
Mean low water level
MZ 38 Steel sheet pile
1V: 1.5 H
1V: 0.58H
Rock dike
76 mm maximum
size
–11.56 m
Fine sand
–18.29 m
0
10 m
Scale
Figure 9.31 Pier C bulkhead—Long Beach harbor (Adapted after Casagrande, 1973)
variation of the lateral earth pressure between May 24, 1949 (the day construction was
completed) and August 6, 1949. On May 24, the lateral earth pressure reached an active
state, as shown in Figure 9.32a, due to the wall yielding. Between May 24 and June 3, the
anchor resisted further yielding and the lateral earth pressure increased to the at-rest state
(Figure 9.32b). However, the flexibility of the sheet piles ultimately resulted in a gradual
decrease in the lateral earth-pressure distribution on the sheet piles (see Figure 9.32c).
May 24
June 3
August 6
+5.18 m
151.11
MN/m2
+5.18 m
177.33
MN/m2
+5.18 m
213.9
MN/m2
–3.05 m
–3.05 m
–11.56 m
–11.56 m
(a)
0
–3.05 m
(b)
100
–11.56 m
(c)
200 kN/m2
Pressure scale
Figure 9.32 Measured stresses at Station 27 ⫹ 30—Pier C bulkhead, Long Beach (Adapted after
Casagrande, 1973)
9.14 Field Observations for Anchor Sheet Pile Walls
481
With time, the stress on the tie rods for the anchor increased as shown in the following table.
Stress on anchor
tie rod (MN/m2)
Date
May 24, 1949
June 3, 1949
June 11, 1949
July 12, 1949
August 6, 1949
151.11
177.33
193.2
203.55
213.9
These observations show that the magnitude of the active earth pressure may vary with
time and depend greatly on the flexibility of the sheet piles. Also, the actual variations in
the lateral earth-pressure diagram may not be identical to those used for design.
Bulkhead—Toledo, Ohio (1961)
A typical cross section of a Toledo bulkhead completed in 1961 is shown in Figure 9.33.
The foundation soil was primarily fine to medium sand, but the dredge line did cut into
highly overconsolidated clay. Figure 9.33 also shows the actual measured values of bending moment along the sheet-pile wall. Casagrande (1973) used the Rankine active earthpressure distribution to calculate the maximum bending moment according to the free
earth support method with and without Rowe’s moment reduction.
Maximum predicted
bending moment, Mmax
Design method
Free earth support method
Free earth support method with Rowe’s moment reduction
Top of fill
0
2
146.5 kN-m
78.6 kN-m
Scale
0
Tie
rod
100
200
kN-m
81 kN-m
4
65 kN-m
May 1961
6
180 kN-m
8
10
205 kN-m
12
14
Depth (m)
Dredge
line
Figure 9.33 Bending
moment from straingage measurements at
test location 3, Toledo
bulkhead (Adapted after
Casagrande, 1973)
482 Chapter 9: Sheet Pile Walls
Comparisons of these magnitudes of Mmax with those actually observed show that the field values are substantially larger. The reason probably is that the backfill was primarily fine sand
and the measured active earth-pressure distribution was larger than that predicted theoretically.
9.15
Free Earth Support Method for Penetration of Clay
Figure 9.34 shows an anchored sheet-pile wall penetrating a clay soil and with a granular
soil backfill. The diagram of pressure distribution above the dredge line is similar to that
shown in Figure 9.12. From Eq. (9.42), the net pressure distribution below the dredge line
(from z 5 L1 1 L2 to z 5 L1 1 L2 1 D) is
s6 5 4c 2 (gL1 1 grL2 )
For static equilibrium, the sum of the forces in the horizontal direction is
P1 2 s6D 5 F
(9.84)
where
P1 5 area of the pressure diagram ACD
F 5 anchor force per unit length of the sheet pile wall
A
l1
L1 O
F
␴1
Water level
l2
C
Sand, ␥, ␾
z
Sand
␥sat, ␾
L2
P1
z1
␴2
Dredge line
D
E
Clay
Clay
␥sat
␾=0
c
D
F
␴6
B
Figure 9.34 Anchored sheet-pile wall penetrating clay
9.15 Free Earth Support Method for Penetration of Clay
483
Again, taking the moment about Or produces
P1 (L1 1 L2 2 l1 2 z1 ) 2 s6D¢l2 1 L2 1
D
≤ 50
2
Simplification yields
s6D2 1 2s6D(L1 1 L2 2 l1 ) 2 2P1 (L1 1 L2 2 l1 2 z1 ) 5 0
(9.85)
Equation (9.85) gives the theoretical depth of penetration, D.
As in Section 9.9, the maximum moment in this case occurs at a depth
L1 , z , L1 1 L2 . The depth of zero shear (and thus the maximum moment) may be
determined from Eq. (9.69).
A moment reduction technique similar to that in Section 9.11 for anchored sheet
piles penetrating into clay has also been developed by Rowe (1952, 1957). This technique
is presented in Figure 9.35, in which the following notation is used:
1. The stability number is
Sn 5 1.25
c
(gL1 1 grL2 )
(9.86)
where c 5 undrained cohesion (f 5 0) .
For the definition of g, gr, L1 , and L2 , see Figure 9.34.
2. The nondimensional wall height is
a5
L1 1 L2
L1 1 L2 1 Dactual
(9.87)
3. The flexibility number is r [see Eq. (9.74)]
4. Md 5 design moment
Mmax 5 maximum theoretical moment
The procedure for moment reduction, using Figure 9.35, is as follows:
Obtain Hr 5 L1 1 L2 1 Dactual .
Determine a 5 (L1 1 L2 )>Hr.
Determine Sn [from Eq. (9.86)].
For the magnitudes of a and Sn obtained in Steps 2 and 3, determine
Md>Mmax for various values of log r from Figure 9.35, and plot Md>Mmax
against log r.
Step 5. Follow Steps 1 through 9 as outlined for the case of moment reduction of
sheet-pile walls penetrating granular soil. (See Section 9.11.)
Step 1.
Step 2.
Step 3.
Step 4.
484 Chapter 9: Sheet Pile Walls
1.0
Log ␳ = –3.1
0.8
Md
Mmax
␣ = 0.8
0.7
0.6
0.6
0.4
1.0
Log ␳ = –2.6
0.8
Md
Mmax
␣ = 0.8
0.6
0.6
0.7
0.4
1.0
Log ␳ = –2.0
0.8
Figure 9.35 Plot of Md>Mmax
against stability number for sheet-pile
wall penetrating
clay (From Rowe, P. W. (1957).
“Sheet Pile Walls in Clay,”
Proceedings, Institute of Civil
Engineers, Vol. 7, pp. 654–692.)
Md
Mmax
␣ = 0.8
0.6
0.7
0.6
0.4
0
0.5
1.0
Stability number, Sn
1.5
1.75
Example 9.10
In Figure 9.34, let L1 5 3 m, L2 5 6 m, and l1 5 1.5 m. Also, let g 5 17 kN>m3,
gsat 5 20 kN>m3, fr 5 35°, and c 5 41 kN>m2.
a. Determine the theoretical depth of embedment of the sheet-pile wall.
b. Calculate the anchor force per unit length of the wall.
Solution
Part a
We have
Ka 5 tan2 ¢ 45 2
fr
35
≤ 5 tan2 ¢45 2 ≤ 5 0.271
2
2
9.15 Free Earth Support Method for Penetration of Clay
485
and
Kp 5 tan2 ¢ 45 1
fr
35
≤ 5 tan2 ¢ 45 1 ≤ 5 3.69
2
2
From the pressure diagram in Figure 9.36,
s1r 5 gL1Ka 5 (17) (3) (0.271) 5 13.82 kN>m2
s2r 5 (gL1 1 grL2 )Ka 5 3(17) (3) 1 (20 2 9.81) (6) 4 (0.271) 5 30.39 kN>m2
P1 5 areas 1 1 2 1 3 5 1>2(3) (13.82) 1 (13.82) (6) 1 1>2(30.39 2 13.82) (6)
5 20.73 1 82.92 1 49.71 5 153.36 kN>m
and
3
6
6
(20.73) ¢ 6 1 ≤ 1 (82.92) ¢ ≤ 1 (49.71) ¢ ≤
3
2
3
z1 5
5 3.2 m
153.36
From Eq. (9.85),
s6D2 1 2s6D(L1 1 L2 2 l1 ) 2 2P1 (L1 1 L2 2 l1 2 z1 ) 5 0
s6 5 4c 2 (gL1 1 grL2 ) 5 (4) (41) 2 3(17) (3)
1 (20 2 9.81) (6)4 5 51.86 kN>m2
So,
(51.86)D2 1 (2) (51.86) (D) (3 1 6 2 1.5)
2 (2) (153.36) (3 1 6 2 1.5 2 3.2) 5 0
l1 1.5 m
L1 3 m
l2 1.5 m
1
␴1 13.82 kN/m2
L2 6 m
2
3
␴2 30.39 kN/m2
1.6 m D
␴ 6 51.86 kN/m2
Figure 9.36 Free earth
support method, sheet pile
penetrating into clay
486 Chapter 9: Sheet Pile Walls
or
D2 1 15D 2 25.43 5 0
Hence,
D < 1.6 m
Part b
From Eq. (9.84),
F 5 P1 2 s6D 5 153.36 2 (51.86) (1.6) 5 70.38 kN , m
9.16
■
Anchors
Sections 9.9 through 9.15 gave an analysis of anchored sheet-pile walls and discussed how
to obtain the force F per unit length of the sheet-pile wall that has to be sustained by the
anchors. The current section covers in more detail the various types of anchor generally used
and the procedures for evaluating their ultimate holding capacities.
The general types of anchor used in sheet-pile walls are as follows:
1.
2.
3.
4.
Anchor plates and beams (deadman)
Tie backs
Vertical anchor piles
Anchor beams supported by batter (compression and tension) piles
Anchor plates and beams are generally made of cast concrete blocks. (See Figure
9.37a.) The anchors are attached to the sheet pile by tie-rods. A wale is placed at the front
or back face of a sheet pile for the purpose of conveniently attaching the tie-rod to the
wall. To protect the tie rod from corrosion, it is generally coated with paint or asphaltic
materials.
In the construction of tiebacks, bars or cables are placed in predrilled holes (see
Figure 9.37b) with concrete grout (cables are commonly high-strength, prestressed steel
tendons). Figures 9.37c and 9.37d show a vertical anchor pile and an anchor beam with
batter piles.
Placement of Anchors
The resistance offered by anchor plates and beams is derived primarily from the passive
force of the soil located in front of them. Figure 9.37a, in which AB is the sheet-pile wall,
shows the best location for maximum efficiency of an anchor plate. If the anchor is placed
inside wedge ABC, which is the Rankine active zone, it would not provide any resistance
to failure. Alternatively, the anchor could be placed in zone CFEH. Note that line DFG is
the slip line for the Rankine passive pressure. If part of the passive wedge is located inside
the active wedge ABC, full passive resistance of the anchor cannot be realized upon
9.16 Anchors
45 ␾/2
A
45 ␾/2
D
Groundwater
table
45 ␾/2
I
Anchor
F
plate
G
H or beam
487
Sheet
pile
C
Anchor
plate
or beam
E
Wale
Tie rod
B
Section
Plan
(a)
45 ␾/2
45 ␾/2
45 ␾/2
Tie rod
Groundwater table
Groundwater table
Tie rod
or cable
Anchor pile
Concrete
grout
(b)
(c)
45 ␾/2
Groundwater
table
Tie rod
Anchor beam
Compression
pile
Tension
pile
(d)
Figure 9.37 Various types of anchoring for sheet-pile walls: (a) anchor plate
or beam; (b) tieback; (c) vertical anchor pile; (d) anchor beam with batter piles
failure of the sheet-pile wall. However, if the anchor is placed in zone ICH, the Rankine
passive zone in front of the anchor slab or plate is located completely outside the Rankine
active zone ABC. In this case, full passive resistance from the anchor can be realized.
Figures 9.37b, 9.37c, and 9.37d also show the proper locations for the placement
of tiebacks, vertical anchor piles, and anchor beams supported by batter piles.
488 Chapter 9: Sheet Pile Walls
9.17
Holding Capacity of Anchor Plates in Sand
Semi-Empirical Method
Ovesen and Stromann (1972) proposed a semi-empirical method for determining the ultimate
resistance of anchors in sand. Their calculations, made in three steps, are carried out as follows:
Step 1. Basic Case. Determine the depth of embedment, H. Assume that the anchor
slab has height H and is continuous (i.e., B 5 length of anchor slab perpendicular to the cross section 5 ` ), as shown in Figure 9.38, in which the
following notation is used:
Pp 5 passive force per unit length of anchor
Pa 5 active force per unit length of anchor
fr 5 effective soil friction angle
dr 5 friction angle between anchor slab and soil
Pult
r 5 ultimate resistance per unit length of anchor
W 5 effective weight per unit length of anchor slab
Also,
r 5 12 gH 2Kp cos d r 2 Pa cos fr 5 12 gH 2Kp cos dr 2 12 gH 2Ka cos fr
Pult
5 12 gH 2 (Kp cos dr 2 Ka cos fr)
(9.88)
where
Ka 5 active pressure coefficient with dr 5 fr
(see Figure 9.39a)
Kp 5 passive pressure coefficient
To obtain Kp cos dr, first calculate
Kp sin dr 5
W 1 Pa sin fr
1
2
2 gH
5
W 1 12gH 2Ka sin fr
45 ␾/2
Pa
H
(9.89)
1
2
2 gH
45 ␾/2
Pult
␾
Sand
␦
Pp
␥
␾
Figure 9.38 Basic case: continuous vertical anchor in granular soil
9.17 Holding Capacity of Anchor Plates in Sand
489
0.7
0.6
0.5
Pa
␾
0.4
Arc of log spiral
Ka
0.3
0.2
0.1
10
40
20
30
Soil friction angle, ␾(deg)
(a)
45
14
12
45
10
40
8
Kp cos ␦
35
6
30
4
␾ = 25
3
2
0
1
2
3
4
5
Kp sin ␦
(b)
Figure 9.39 (a) Variation of Ka for dr 5 fr, (b) variation of Kp cos dr
with Kp sin d r (Based on Ovesen and Stromann, 1972)
Then use the magnitude of Kp sin dr obtained from Eq. (9.89) to estimate
the magnitude of Kp cos dr from the plots given in Figure 9.39b.
Step 2. Strip Case. Determine the actual height h of the anchor to be constructed. If
a continuous anchor (i.e., an anchor for which B 5 ` ) of height h is placed
in the soil so that its depth of embedment is H, as shown in Figure 9.40, the
ultimate resistance per unit length is
490 Chapter 9: Sheet Pile Walls
Sand
␥
␾
H
h
Pus
Figure 9.40 Strip case: vertical anchor
Pus
r 5 C
Cov 1 1
H
Cov 1 ¢ ≤
h
S
Pult
r
c
(9.90)
Eq. 9.88
where
P rus 5 ultimate resistance for the strip case
Cov 5 19 for dense sand and 14 for loose sand
Step 3. Actual Case. In practice, the anchor plates are placed in a row with centerto-center spacing Sr, as shown in Figure 9.41a. The ultimate resistance of
each anchor is
Sand
␥
␾
B
H
S
S
h
(a)
0.5
Dense sand
(Be – B)/(H + h)
0.4
0.3
Loose sand
0.2
0.1
0
0
0.5
(S – B)/(H – h)
(b)
1.0
1.25
Figure 9.41 (a) Actual
case for row of anchors;
(b) variation of
(Be 2 B)> (H 1 h) with
(Sr 2 B)> (H 1 h)
(Based on Ovesen and
Stromann, 1972)
9.17 Holding Capacity of Anchor Plates in Sand
Pult 5 Pus
r Be
491
(9.91)
where Be 5 equivalent length.
The equivalent length is a function of Sr, B, H, and h. Figure 9.41b shows a
plot of (Be 2 B)> (H 1 h) against (Sr 2 B)> (H 1 h) for the cases of
loose and dense sand. With known values of Sr, B, H, and h, the value of Be
can be calculated and used in Eq. (9.91) to obtain Pult .
Stress Characteristic Solution
Neely, Stuart, and Graham (1973) proposed a stress characteristic solution for anchor pullout resistance using the equivalent free surface concept. Figure 9.42 shows the assumed
failure surface for a strip anchor. In this figure, OX is the equivalent free surface. The shear
stress (so) mobilized along OX can be given as
m5
so
9
so tan
f9
(9.92)
where
m ⫽ shear stress mobilization factor
␴o⬘ ⫽ effective normal stress along OX
Using this analysis, the ultimate resistance (Pult) of an anchor (length ⫽ B and
height ⫽ h) can be given as
Pult ⫽ M␥q (␥h2)BFs
(9.93)
where
M␥q ⫽ force coefficient
Fs ⫽ shape factor
␥ ⫽ effective unit weight of soil
The variations of M␥q for m ⫽ 0 and 1 are shown in Figure 9.43. For conservative
design, M␥q with m ⫽ 0 may be used. The shape factor (Fs) determined experimentally is
shown in Figure 9.44 as a function of B/h and H/h.
X
␴o
so
␥
␾
O
H
Pult
h
m=
so
␴o tan ␾
Figure 9.42 Assumed failure surface in soil for
stress characteristic solution
492 Chapter 9: Sheet Pile Walls
200
100
M␥q (log scale)
50
20
␾= 45°
10
␾ = 40°
␾ = 35°
5
m=0
m=1
␾ = 30°
2
1
0
1
2
4
3
5
H/h
Figure 9.43 Variation of M␥q with
H/h and ␾⬘ (After Neeley et al., 1973.
With permission from ASCE.)
2.5
B/h = 1.0
Shape factor, Fs
2.0
1.5
2.0
2.75
3.5
5.0
1.0
0.5
0
1
2
3
H/h
4
5
Figure 9.44 Variation of shape factor with
H/h and B/h (After Neeley et al., 1973. With
permission from ASCE.)
Empirical Correlation Based on Model Tests
Ghaly (1997) used the results of 104 laboratory tests, 15 centrifugal model tests, and 9
field tests to propose an empirical correlation for the ultimate resistance of single anchors.
The correlation can be written as
Pult 5
5.4 H 2 0.28
¢
≤ gAH
tan fr A
where A 5 area of the anchor 5 Bh.
(9.94)
9.17 Holding Capacity of Anchor Plates in Sand
493
Ghaly also used the model test results of Das and Seeley (1975) to develop a
load–displacement relationship for single anchors. The relationship can be given as
P
u 0.3
5 2.2¢ ≤
Pult
H
(9.95)
where u 5 horizontal displacement of the anchor at a load level P.
Equations (9.94) and (9.95) apply to single anchors (i.e., anchors for which Sr>B 5 `).
For all practical purposes, when Sr>B < 2 the anchors behave as single anchors.
Factor of Safety for Anchor Plates
The allowable resistance per anchor plate may be given as
Pall 5
Pult
FS
where FS 5 factor of safety.
Generally, a factor of safety of 2 is suggested when the method of Ovesen and
Stromann is used. A factor of safety of 3 is suggested for Pult calculated by Eq. (9.94).
Spacing of Anchor Plates
The center-to-center spacing of anchors, Sr, may be obtained from
Sr 5
Pall
F
where F 5 force per unit length of the sheet pile.
Example 9.11
Refer to Figure 9.41a. Given: B ⫽ h ⫽ 0.4 m, S⬘ ⫽ 1.2 m, H ⫽ 1 m, ␥ ⫽ 16.51 kN/m3,
and ␾⬘ ⫽ 35°. Determine the ultimate resistance for each anchor plate. The anchor
plates are made of concrete and have thicknesses of 0.15 m.
Solution
From Figure 9.39a for ␾⬘ ⫽ 35°, the magnitude of Ka is about 0.26.
W ⫽ Ht␥concrete ⫽ (1 m)(0.15 m)(23.5 kN/m3)
⫽ 3.525 kN/m
From Eq. (9.89),
Kp sin d9 5
5
W 1 1> 2gH 2Ka sin f9
> 2gH 2
3.525 1 (0.5) (16.51) (1) 2 (0.26) (sin 35)
1
(0.5) (16.51) (1) 2
5 0.576
494 Chapter 9: Sheet Pile Walls
From Figure 9.39b with ␾⬘ ⫽ 35° and Kp sin ␦⬘ ⫽ 0.576, the value of Kp cos ␦⬘ is
about 4.5. Now, using Eq. (9.88),
Pult
⬘ ⫽ 21␥H2(Kp cos ␦⬘ ⫺ Ka cos ␾⬘)
⫽ (12)(16.51)(1)2[4.5 ⫺ (0.26)(cos 35)] ⫽ 35.39 kN/m
In order to calculate Pus
⬘, let us assume the sand to be loose. So, Cov in Eq. (9.90) is
equal to 14. Hence,
Pus
r 5D
Cov 1 1
H
Cov 1 a b
h
T Pult
r 5D
14 1 1
T 5 32.17 kN>m
1
14 1 a b
0.4
Sr 2 B
1.2 2 0.4
0.8
5
5
5 0.571
H1h
1 1 0.4
1.4
For (S⬘ ⫺ B)/(H ⫹ h) ⫽ 0.571 and loose sand, Figure 9.41b yields
Be 2 B
5 0.229
H2h
So
Be ⫽ (0.229)(H ⫹ h) ⫹ B ⫽ (0.229)(1 ⫹ 0.4) ⫹ 0.4
⫽ 0.72
Hence, from Eq. (9.91)
Pult ⫽ Pus
⬘ Be ⫽ (32.17)(0.72) ⫽ 23.16 kN
■
Example 9.12
Refer to a single anchor given in Example 9.11 using the stress characteristic solution.
Estimate the ultimate anchor resistance. Use m ⫽ 0 in Figure 9.43.
Solution
Given: B ⫽ h ⫽ 0.4 m and H ⫽ 1 m.
Thus,
H
1m
5
5 2.5
h
0.4 m
0.4 m
B
5
51
h
0.4 m
9.19 Ultimate Resistance of Tiebacks
495
From Eq. (9.93),
Pult ⫽ M␥q␥h2 BFs
From Figure 9.43, with ␾⬘ ⫽ 35° and H/h ⫽ 2.5, M␥q ⬇ 18.2. Also, from Figure 9.44,
with H/h ⫽ 2.5 and B/h ⫽ 1, Fs ⬇ 1.8. Hence,
Pult ⫽ (18.2)(16.51)(0.4)2(0.4)(1.8) ⬇ 34.62 kN
■
Example 9.13
Solve Example Problem 9.12 using Eq. (9.94).
Solution
From Eq. (9.94),
Pult 5
5.4 H 2 0.28
a
b gAH
tan f9 A
H⫽1m
A ⫽ Bh ⫽ (0.4 ⫻ 0.4) ⫽ 0.16 m2
Pult 5
9.18
5.4 (1) 2 0.28
c
d (16.51) (0.16) (1) < 34.03 kN
tan 35 0.16
■
Holding Capacity of Anchor Plates
in Clay (f 5 0 Condition)
Relatively few studies have been conducted on the ultimate resistance of anchor plates in
clayey soils (f 5 0). Mackenzie (1955) and Tschebotarioff (1973) identified the nature of
variation of the ultimate resistance of strip anchors and beams as a function of H, h, and c
(undrained cohesion based on f 5 0) in a nondimensional form based on laboratory
model test results. This is shown in the form of a nondimensional plot in Figure 9.45
(Pult>hBc versus H>h) and can be used to estimate the ultimate resistance of anchor plates
in saturated clay (f 5 0).
9.19
Ultimate Resistance of Tiebacks
According to Figure 9.46, the ultimate resistance offered by a tieback in sand is
Pult 5 pdl sor K tan fr
(9.96)
Chapter 9: Sheet Pile Walls
12
10
8
Pult
hBc
496
6
4
2
0
0
5
10
15
20
H
h
Pult
with H>h for plate anchors in clay
hBc
(Based on Mackenzie (1955) and Tschebotarioff (1973))
Figure 9.45 Experimental variation of
z
l
d
Figure 9.46 Parameters for defining the
ultimate resistance of tiebacks
where
fr 5 effective angle of friction of soil
sor 5 average effective vertical stress (5gz in dry sand)
K 5 earth pressure coefficient
The magnitude of K can be taken to be equal to the earth pressure coefficient at rest (Ko )
if the concrete grout is placed under pressure (Littlejohn, 1970). The lower limit of K can
be taken to be equal to the Rankine active earth pressure coefficient.
In clays, the ultimate resistance of tiebacks may be approximated as
Pult 5 pdlca
where ca 5 adhesion.
(9.97)
Problems
497
The value of ca may be approximated as 23cu (where cu 5 undrained cohesion). A
factor of safety of 1.5 to 2 may be used over the ultimate resistance to obtain the allowable
resistance offered by each tieback.
Problems
9.1
9.2
9.3
9.4
9.5
9.6
Figure P9.1 shows a cantilever sheet pile wall penetrating a granular soil. Here,
L1 5 4 m, L2 5 8 m, g 5 16.1 kN>m3, gsat 5 18.2 kN>m3, and f r 5 32°.
a. What is the theoretical depth of embedment, D?
b. For a 30% increase in D, what should be the total length of the sheet piles?
c. Determine the theoretical maximum moment of the sheet pile.
Redo Problem 9.1 with the following: L1 5 3 m, L2 5 6 m, g 5 17.3 kN>m3,
gsat 5 19.4 kN>m3 , and fr 5 30°.
Refer to Figure 9.10. Given: L 5 3 m, g 5 16.7 kN>m3, and fr 5 30°. Calculate
the theoretical depth of penetration, D, and the maximum moment.
Refer to Figure P9.4, for which L1 5 2.4 m, L2 5 4.6 m, g 5 15.7 kN>m3 ,
gsat 5 17.3 kN>m3 , and f r 5 30°, and c 5 29 kN>m2.
a. What is the theoretical depth of embedment, D?
b. Increase D by 40%. What length of sheet piles is needed?
c. Determine the theoretical maximum moment in the sheet pile.
Refer to Figure 9.14. Given: L 5 4 m ; for sand, g 5 16 kN>m3; fr 5 35°; and,
for clay, gsat 5 19.2 kN>m3 and c 5 45 kN>m2. Determine the theoretical value
of D and the maximum moment.
An anchored sheet pile bulkhead is shown in Figure P9.6. Let L1 5 4 m,
L2 5 9 m, l1 5 2 m , g 5 17 kN>m3, gsat 5 19 kN>m3 , and fr 5 34°.
a. Calculate the theoretical value of the depth of embedment, D.
b. Draw the pressure distribution diagram.
c. Determine the anchor force per unit length of the wall.
Use the free earth-support method.
L1
Sand
␥
c 0
␾
L2
Sand
␥sat
c 0
␾
D
Sand
␥sat
c 0
␾
Water table
Dredge line
Figure P9.1
498 Chapter 9: Sheet Pile Walls
L1
Sand
␥
c 0
␾
L2
Sand
␥sat
c 0
␾
Water table
Clay
c
␾ 0
D
Figure P9.4
l1
Water table
L1
Anchor
Sand
c 0
␥
␾
L2
Sand
␥sat
c 0
␾
D
Sand
␥sat
c 0
␾
Figure P9.6
9.7
9.8
9.9
In Problem 9.6, assume that Dactual 5 1.3Dtheory .
a. Determining the theoretical maximum moment.
b. Using Rowe’s moment reduction technique, choose a sheet pile section. Take
E 5 210 3 103 MN>m2 and sall 5 210,000 kN>m2.
Refer to Figure P9.6. Given: L1 5 4 m, L2 5 8 m, l1 5 l2 5 2 m, ␥ 5 16 kN/m3,
␥sat 5 18.5 kN/m3, and fr 5 35°. Use the charts presented in Section 9.10 and determine:
a. Theoretical depth of penetration
b. Anchor force per unit length
c. Maximum moment in the sheet pile.
Refer to Figure P9.6, for which L1 5 4 m, L2 5 7 m, l1 5 1.5 m, g 5 18 kN>m3,
gsat 5 19.5 kN>m3, and fr 5 30°. Use the computational diagram method
(Section 9.12) to determine D, F, and Mmax. Assume that C 5 0.68 and R 5 0.6.
Problems
499
9.10 An anchored sheet-pile bulkhead is shown in Figure P9.10. Let
L1 5 2 m, L2 5 6 m, l1 5 1 m, g 5 16 kN>m3, gsat 5 18.86 kN>m3, fr 5 32°,
and c 5 27 kN>m2.
a. Determine the theoretical depth of embedment, D.
b. Calculate the anchor force per unit length of the sheet-pile wall.
Use the free earth support method.
9.11 In Figure 9.41a, for the anchor slab in sand, H 5 1.52 m, h 5 0.91 m,
B 5 1.22 m, S r 5 2.13 m, fr 5 30°, and g 5 17.3 kN>m3. The anchor
plates are made of concrete and have a thickness of 76 mm. Using Ovesen
and Stromann’s method, calculate the ultimate holding capacity of each anchor.
Take gconcrete 5 23.58 kN>m3.
9.12 A single anchor slab is shown in Figure P9.12. Here, H 5 0.9 m, h 5 0.3 m,
g 5 17 kN>m3, and fr 5 32°. Calculate the ultimate holding capacity of the anchor slab if the width B is (a) 0.3 m, (b) 0.6 m, and (c) 5 0.9 m.
(Note: center-to-center spacing, S r 5 ` .) Use the empirical correlation given in
Section 9.17 [Eq. (9.94)].
9.13 Repeat Problem 9.12 using Eq. (9.93). Use m 5 0 in Figure 9.43.
l1
Anchor
L1
Water table
Sand
c 0
␥
␾
L2
Sand
␥sat
c 0
␾
D
Clay
c
␾ 0
Figure P9.10
␥
c 0
␾
H
h
Pult
Figure P9.12
500 Chapter 9: Sheet Pile Walls
References
BLUM, H. (1931) Einspannungsverhältnisse bei Bohlwerken, W. Ernst und Sohn, Berlin, Germany.
CASAGRANDE, L. (1973). “Comments on Conventional Design of Retaining Structures,” Journal of
the Soil Mechanics and Foundations Division, ASCE, Vol. 99, No. SM2, pp. 181–198.
CORNFIELD, G. M. (1975). “Sheet Pile Structures,” in Foundation Engineering Handbook, ed. H. F.
Wintercorn and H. Y. Fang, Van Nostrand Reinhold, New York, pp. 418–444.
DAS, B. M., and SEELEY, G. R. (1975). “Load–Displacement Relationships for Vertical Anchor
Plates,” Journal of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 101, No, GT7, pp. 711–715.
GHALY, A. M. (1997). “Load–Displacement Prediction for Horizontally Loaded Vertical Plates.”
Journal of Geotechnical and Geoenvironmental Engineering, ASCE, Vol. 123, No. 1,
pp. 74–76.
HAGERTY, D. J., and NOFAL, M. M. (1992). “Design Aids: Anchored Bulkheads in Sand,” Canadian
Geotechnical Journal, Vol. 29, No. 5, pp. 789–795.
LITTLEJOHN, G. S. (1970). “Soil Anchors,” Proceedings, Conference on Ground Engineering, Institute of Civil Engineers, London, pp. 33–44.
MACKENZIE, T. R. (1955). Strength of Deadman Anchors in Clay, M.S. Thesis, Princeton University,
Princeton, N. J.
NATARAJ, M. S., and HOADLEY, P. G. (1984). “Design of Anchored Bulkheads in Sand,” Journal of
Geotechnical Engineering, American Society of Civil Engineers, Vol. 110, No. GT4,
pp. 505–515.
NEELEY, W. J., STUART, J. G., and GRAHAM, J. (1973). “Failure Loads of Vertical Anchor Plates is
Sand,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil
Engineers, Vol. 99, No. SM9, pp. 669–685.
OVESEN, N. K., and STROMANN, H. (1972). “Design Methods for Vertical Anchor Slabs in Sand,” Proceedings, Specialty Conference on Performance of Earth and Earth-Supported Structures.
American Society of Civil Engineers, Vol. 2.1, pp. 1481–1500.
ROWE, P. W. (1952). “Anchored Sheet Pile Walls,” Proceedings, Institute of Civil Engineers, Vol. 1, Part
1, pp. 27–70.
ROWE, P. W. (1957). “Sheet Pile Walls in Clay,” Proceedings, Institute of Civil Engineers, Vol. 7,
pp. 654 –692.
TSCHEBOTARIOFF, G. P. (1973). Foundations, Retaining and Earth Structures, 2nd ed., McGraw-Hill,
New York.
TSINKER, G. P. (1983). “Anchored Street Pile Bulkheads: Design Practice,” Journal of Geotechnical
Engineering, American Society of Civil Engineers, Vol. 109, No. GT8, pp. 1021–1038.
10
10.1
Braced Cuts
Introduction
Sometimes construction work requires ground excavations with vertical or near-vertical
faces—for example, basements of buildings in developed areas or underground transportation facilities at shallow depths below the ground surface (a cut-and-cover type of
construction). The vertical faces of the cuts need to be protected by temporary bracing systems to avoid failure that may be accompanied by considerable settlement or by bearing
capacity failure of nearby foundations.
Figure 10.1 shows two types of braced cut commonly used in construction work.
One type uses the soldier beam (Figure 10.1a), which is driven into the ground before
excavation and is a vertical steel or timber beam. Laggings, which are horizontal timber planks, are placed between soldier beams as the excavation proceeds. When the
excavation reaches the desired depth, wales and struts (horizontal steel beams) are
installed. The struts are compression members. Figure 10.1b shows another type of
braced excavation. In this case, interlocking sheet piles are driven into the soil before
excavation. Wales and struts are inserted immediately after excavation reaches the
appropriate depth.
Figure 10.2 shows the braced-cut construction used for the Chicago subway in
1940. Timber lagging, timber struts, and steel wales were used. Figure 10.3 shows a
braced cut made during the construction of the Washington, DC, metro in 1974. In this
cut, timber lagging, steel H-soldier piles, steel wales, and pipe struts were used.
To design braced excavations (i.e., to select wales, struts, sheet piles, and soldier
beams), an engineer must estimate the lateral earth pressure to which the braced cuts will
be subjected. The theoretical aspects of the lateral earth pressure on a braced cut were discussed in Section 7.8. The total active force per unit length of the wall (Pa ) was calculated
using the general wedge theory. However, that analysis does not provide the relationships
required for estimating the variation of lateral pressure with depth, which is a function of
several factors, such as the type of soil, the experience of the construction crew, the type
of construction equipment used, and so forth. For that reason, empirical pressure envelopes
developed from field observations are used for the design of braced cuts. This procedure
is discussed in the next section.
501
502 Chapter 10: Braced Cuts
Wale
Strut
Strut
A
A
Soldier
beam
Lagging
Lagging
Wale
Wedge
Elevation
Plan
(a)
Wale
Strut
Strut
A
A
Sheet
pile
Wale
Plan
Elevation
(b)
Figure 10.1 Types of braced cut: (a) use of soldier beams; (b) use of sheet piles
10.2
Pressure Envelope for Braced-Cut Design
As mentioned in Section 10.1, the lateral earth pressure in a braced cut is dependent on the
type of soil, construction method, and type of equipment used. The lateral earth pressure
changes from place to place. Each strut should also be designed for the maximum load to
10.2 Pressure Envelope for Braced-Cut Design
503
Figure 10.2 Braced cut in Chicago Subway construction, January 1940 (Courtesy of Ralph B.
Peck)
which it may be subjected. Therefore, the braced cuts should be designed using apparentpressure diagrams that are envelopes of all the pressure diagrams determined from measured strut loads in the field. Figure 10.4 shows the method for obtaining the
apparent-pressure diagram at a section from strut loads. In this figure, let P1 , P2 , P3 , P4 , c
be the measured strut loads. The apparent horizontal pressure can then be calculated as
P1
s1 5
d2
(s) ¢ d1 1 ≤
2
P2
s2 5
d2
d3
(s) ¢ 1 ≤
2
2
P3
s3 5
d3
d4
(s) ¢ 1 ≤
2
2
P4
s4 5
d4
d5
(s) ¢ 1 ≤
2
2
504 Chapter 10: Braced Cuts
Image not available due to copyright restrictions
d1
d1
1
P1
d2/2
d2
d2/2
2
P2
d3/2
d3
d3/2
P3
3
d4/2
d4
d4/2
P4
d5
d5/2
d5/2
4
Figure 10.4 Procedure for calculating
apparent-pressure diagram from measured
strut loads
10.2 Pressure Envelope for Braced-Cut Design
505
where
s1 , s2 , s3 , s4 5 apparent pressures
s 5 center-to-center spacing of the struts
Using the procedure just described for strut loads observed from the Berlin subway
cut, Munich subway cut, and New York subway cut, Peck (1969) provided the envelope of
apparent-lateral-pressure diagrams for design of cuts in sand. This envelope is illustrated
in Figure 10.5, in which
sa 5 0.65gHKa
(10.1)
where
g 5 unit weight
H 5 height of the cut
Ka 5 Rankine active pressure coefficient 5 tan2 (45 2 fr>2)
fr 5 effective friction angle of sand
Cuts in Clay
In a similar manner, Peck (1969) also provided the envelopes of apparent-lateral-pressure
diagrams for cuts in soft to medium clay and in stiff clay. The pressure envelope for soft
to medium clay is shown in Figure 10.6 and is applicable to the condition
gH
.4
c
where c 5 undrained cohesion (f 5 0).
The pressure, sa , is the larger of
sa 5 gHB1 2 ¢
4c
≤R
gH
and
sa 5 0.3gH
(10.2)
where g 5 unit weight of clay.
The pressure envelope for cuts in stiff clay is shown in Figure 10.7, in which
sa 5 0.2gH to 0.4gH
is applicable to the condition gH>c < 4.
(with an average of 0.3gH)
(10.3)
506 Chapter 10: Braced Cuts
0.25 H
0.25 H
a
H
a
0.5 H
a
0.75 H
0.25 H
Figure 10.5 Peck’s (1969)
apparent-pressure envelope
for cuts in sand
Figure 10.6 Peck’s (1969)
apparent-pressure envelope for
cuts in soft to medium clay
Figure 10.7 Peck’s (1969)
apparent-pressure envelope
for cuts in stiff clay
When using the pressure envelopes just described, keep the following points in mind:
1.
2.
3.
4.
10.3
They apply to excavations having depths greater than about 6 m.
They are based on the assumption that the water table is below the bottom of the cut.
Sand is assumed to be drained with zero pore water pressure.
Clay is assumed to be undrained and pore water pressure is not considered.
Pressure Envelope for Cuts in Layered Soil
Sometimes, layers of both sand and clay are encountered when a braced cut is being constructed. In this case, Peck (1943) proposed that an equivalent value of cohesion (f 5 0)
should be determined according to the formula (see Figure 10.8a).
cav 5
1
3gsKsH 2s tan fsr 1 (H 2 Hs )nrqu4
2H
(10.4)
where
H 5 total height of the cut
gs 5 unit weight of sand
Hs 5 height of the sand layer
Ks 5 a lateral earth pressure coefficient for the sand layer (<1)
fsr 5 effective angle of friction of sand
qu 5 unconfined compression strength of clay
nr 5 a coefficient of progressive failure (ranging from 0.5 to 1.0; average value 0.75)
10.4 Design of Various Components of a Braced Cut
Sand
s
Hs
H
H1
Clay
1, c1
H2
Clay
2, c2
Hn
Clay
n , cn
507
H
Clay
c
Hc
(a)
(b)
Figure 10.8 Layered soils in braced cuts
The average unit weight of the layers may be expressed as
ga 5
1
3g H 1 (H 2 Hs )gc4
H s s
(10.5)
where gc 5 saturated unit weight of clay layer.
Once the average values of cohesion and unit weight are determined, the pressure
envelopes in clay can be used to design the cuts.
Similarly, when several clay layers are encountered in the cut (Figure 10.8b), the
average undrained cohesion becomes
cav 5
1
(c H 1 c2H2 1 c 1 cnHn )
H 1 1
(10.6)
where
c1 , c2 , c, cn 5 undrained cohesion in layers 1, 2, c, n
H1 , H2 , c, Hn 5 thickness of layers 1, 2, c, n
The average unit weight is now
ga 5
10.4
1
(g H 1 g2H2 1 g3H3 1 c 1 gnHn )
H 1 1
(10.7)
Design of Various Components of a Braced Cut
Struts
In construction work, struts should have a minimum vertical spacing of about 2.75 m
or more. Struts are horizontal columns subject to bending. The load-carrying capacity
of columns depends on their slenderness ratio, which can be reduced by providing
508 Chapter 10: Braced Cuts
vertical and horizontal supports at intermediate points. For wide cuts, splicing the
struts may be necessary. For braced cuts in clayey soils, the depth of the first
strut below the ground surface should be less than the depth of tensile crack, zc . From
Eq. (7.8),
sar 5 gzKa 2 2cr"Ka
where Ka 5 coefficient of Rankine active pressure.
For determining the depth of tensile crack,
sar 5 0 5 gzcKa 2 2cr"Ka
or
zc 5
2cr
"Kag
With f 5 0, Ka 5 tan2 (45 2 f>2) 5 1, so
zc 5
2c
g
A simplified conservative procedure may be used to determine the strut loads.
Although this procedure will vary, depending on the engineers involved in the project, the
following is a step-by-step outline of the general methodology (see Figure 10.9):
Step 1.
Step 2.
Step 3.
Draw the pressure envelope for the braced cut. (See Figures 10.5, 10.6,
and 10.7.) Also, show the proposed strut levels. Figure 10.9a shows a
pressure envelope for a sandy soil; however, it could also be for a clay.
The strut levels are marked A, B, C, and D. The sheet piles (or soldier
beams) are assumed to be hinged at the strut levels, except for the top and
bottom ones. In Figure 10.9a, the hinges are at the level of struts B and C.
(Many designers also assume the sheet piles or soldier beams to be hinged
at all strut levels except for the top.)
Determine the reactions for the two simple cantilever beams (top and bottom) and all the simple beams between. In Figure 10.9b, these reactions are
A, B1 , B2 , C1 , C2 , and D.
The strut loads in the figure may be calculated via the formulas
PA 5 (A) (s)
PB 5 (B1 1 B2 ) (s)
PC 5 (C1 1 C2 ) (s)
and
PD 5 (D) (s)
(10.8)
10.4 Design of Various Components of a Braced Cut
509
d1
A
d2
B
Hinges
a
d3
C
Simple
cantilever
d1
d4
Section
A
D
a
d2
d5
B1
Simple
beam
B2
a
d3
s
C1
Simple
cantilever
C2
d4
Plan
a
D
d5
(a)
(b)
Figure 10.9 Determination of strut loads: (a) section and plan of the cut; (b) method
for determining strut loads
where
PA , PB , PC , PD 5 loads to be taken by the individual struts at levels
A, B, C, and D, respectively
A, B1 , B2 , C1 , C2 , D 5 reactions calculated in Step 2 (note the unit:
force>unit length of the braced cut)
s 5 horizontal spacing of the struts (see plan in
Figure 10.9a)
Step 4. Knowing the strut loads at each level and the intermediate bracing conditions allows selection of the proper sections from the steel construction
manual.
510 Chapter 10: Braced Cuts
Sheet Piles
The following steps are involved in designing the sheet piles:
Step 1.
Step 2.
Step 3.
For each of the sections shown in Figure 10.9b, determine the maximum
bending moment.
Determine the maximum value of the maximum bending moments (Mmax )
obtained in Step 1. Note that the unit of this moment will be, for example,
kN-m>m length of the wall.
Obtain the required section modulus of the sheet piles, namely,
S5
Step 4.
Mmax
sall
(10.9)
where sall 5 allowable flexural stress of the sheet pile material.
Choose a sheet pile having a section modulus greater than or equal to the
required section modulus from a table such as Table 9.1.
Wales
Wales may be treated as continuous horizontal members if they are spliced properly.
Conservatively, they may also be treated as though they are pinned at the struts. For the
section shown in Figure 10.9a, the maximum moments for the wales (assuming that they
are pinned at the struts) are,
At level A,
Mmax 5
(A) (s2 )
8
At level B,
Mmax 5
(B1 1 B2 )s2
8
At level C,
Mmax 5
(C1 1 C2 )s2
8
and
At level D, Mmax 5
(D) (s2 )
8
where A, B1 , B2 , C1 , C2 , and D are the reactions under the struts per unit length of the wall
(see Step 2 of strut design).
Now determine the section modulus of the wales:
S5
Mmax
sall
The wales are sometimes fastened to the sheet piles at points that satisfy the lateral support
requirements.
10.4 Design of Various Components of a Braced Cut
511
Example 10.1
The cross section of a long braced cut is shown in Figure 10.10a.
a.
b.
c.
d.
Draw the earth-pressure envelope.
Determine the strut loads at levels A, B, and C.
Determine the section modulus of the sheet pile section required.
Determine a design section modulus for the wales at level B.
(Note: The struts are placed at 3 m, center to center, in the plan.) Use
sall 5 170 3 103 kN>m2
Solution
Part a
We are given that g 5 18 kN>m2, c 5 35 kN>m2, and H 5 7 m. So,
(18) (7)
gH
5
5 3.6 , 4
c
35
Thus, the pressure envelope will be like the one in Figure 10.7. The envelope is
plotted in Figure 10.10a with maximum pressure intensity, sa , equal to
0.3gH 5 0.3(18) (7) 5 37.8 kN , m2.
Part b
To calculate the strut loads, examine Figure 10.10b. Taking the moment about B1 , we
have S MB1 5 0, and
1.75
1.75
1
≤ 2 (1.75) (37.8) ¢
≤ 50
A(2.5) 2 ¢ ≤ (37.8) (1.75) ¢ 1.75 1
2
3
2
or
A 5 54.02 kN>m
Also, S vertical forces 5 0. Thus,
1
2 (1.75) (37.8)
1 (37.8) (1.75) 5 A 1 B1
or
33.08 1 66.15 2 A 5 B1
So,
B1 5 45.2 kN>m
Due to symmetry,
B2 5 45.2 kN>m
and
C 5 54.02 kN>m
512 Chapter 10: Braced Cuts
6m
1m
A
1.75 m
Sheet pile
2.5 m
B
3.5 m
37.8
kN/m2
Clay
18 kN/m3
c 35 kN/m2
0
2.5 m
C
1.75 m
1m
(a) Cross section
1.75 m
1.75 m
1.75 m
37.8 kN/m2
37.8 kN/m2
1.75 m
21.6
1m
2.5 m
2.5 m
A
B1
1m
C
B2
(b) Determination of reaction
43.23 kN
43.23 kN
x 1.196 m
B1
A
B2
E
F
10.8 kN
C
10.8 kN
45.2 kN
45.2 kN
(c) Shear force diagram
Figure 10.10 Analysis of a braced cut
10.4 Design of Various Components of a Braced Cut
513
Hence, the strut loads at the levels indicated by the subscripts are
PA 5 54.02 3 horizontal spacing, s 5 54.02 3 3 5 162.06 kN
PB 5 (B1 1 B2 )3 5 (45.2 1 45.2)3 5 271.2 kN
and
PC 5 54.02 3 3 5 162.06 kN
Part c
At the left side of Figure 10.10b, for the maximum moment, the shear force should be
zero. The nature of the variation of the shear force is shown in Figure 10.10c. The location of point E can be given as
x5
reaction at B1
45.2
5
5 1.196 m
37.8
37.8
Also,
37.8
1
1
3 1≤ ¢ ≤
Magnitude of moment at A 5 (1) ¢
2
1.75
3
5 3.6 kN-m>meter of wall
and
Magnitude of moment at E 5 (45.2 3 1.196) 2 (37.8 3 1.196) ¢
1.196
≤
2
5 54.06 2 27.03 5 27.03 kN-m>meter of wall
Because the loading on the left and right sections of Figure 10.10b are the same,
the magnitudes of the moments at F and C (see Figure 10.10c) will be the same as
those at E and A, respectively. Hence, the maximum moment is 27.03 kN-m>meter
of wall.
The section modulus of the sheet piles is thus
S5
Mmax
27.03 kN-m
5
5 15.9 3 1025m3 , m of the wall
sall
170 3 103 kN>m2
Part d
The reaction at level B has been calculated in part b. Hence,
Mmax 5
(B1 1 B2 )s2
(45.2 1 45.2)32
5
5 101.7 kN-m
8
8
and
Section modulus, S 5
101.7
101.7
5
sall
(170 3 1000)
5 0.598 3 1023 m3
■
514 Chapter 10: Braced Cuts
Example 10.2
Refer to the braced cut shown in Figure 10.11, for which ␥ ⫽ 17 kN/m3, ␾⬘ ⫽ 35°,
and cr ⫽ 0. The struts are located 4 m on center in the plan. Draw the earth-pressure
envelope and determine the strut loads at levels A, B, and C.
Solution
For this case, the earth-pressure envelope shown in Figure 10.5 is applicable. Hence,
Ka 5 tan2 a45 2
fr
35
b 5 tan2 a45 2 b 5 0.271
2
2
From Equation (10.1)
sa 5 0.65 gHKa 5 (0.65) (17) (9) (0.271) 5 26.95 kN>m2
Figure 10.12a shows the pressure envelope. Refer to Figure 10.12b and calculate B1:
a MB1 5 0
5
(26.95) (5) a b
2
A5
5 112.29 kN>m
3
B1 5 (26.95) (5) 2 112.29 5 22.46 kN>m
Now, refer to Figure 10.12c and calculate B2:
a MB2 5 0
5m
2m
A
3m
B
Sand
c0
3m
C
1m
Figure 10.11
10.5 Case Studies of Braced Cuts
515
2m
A
3m
B
a 0.65HKa
26.95 kN/m2
3m
C
1m
(a)
26.95
kN/m2
2m
26.95
kN/m2
3m
3m
B1
A
(b)
B2
1m
C
(c)
Figure 10.12 Load diagrams
4
(26.95) (4) a b
2
C5
5 71.87 kN>m
3
B2 5 (26.95) (4) 2 71.87 5 35.93 kN>m
The strut loads are
At A, (112.29)(spacing) ⫽ (112.29)(4) ⫽ 449.16 kN
At B, (B1 ⫹ B2)(spacing) ⫽ (22.46 ⫹ 35.93)(4) ⫽ 233.56 kN
At C, (71.87)(spacing) ⫽ (71.87)(4) ⫽ 287.48 kN
10.5
■
Case Studies of Braced Cuts
The procedure for determining strut loads and the design of sheet piles and wales presented in the preceding sections appears to be fairly straightforward. It is, however, only
possible if a proper pressure envelope is chosen for the design, which is difficult. This section describes some case studies of braced cuts and highlights the difficulties and degree
of judgment needed for successful completion of various projects.
Subway Extension of the Massachusetts Bay Transportation
Authority (MBTA)
Lambe (1970) provided data on the performance of three excavations for the subway
extension of the MBTA in Boston (test sections A, B, and D), all of which were well instrumented. Figure 10.13 gives the details of test section B, where the cut was 17.68 m, including subsoil conditions. The subsoil consisted of gravel, sand, silt, and clay (fill) to a depth
of about 7.93 m, followed by a light gray, slightly organic silt to a depth of 14.02 m. A
layer of coarse sand and gravel with some clay was present from 14.02 m to 16.46 m below
the ground surface. Rock was encountered below 16.46 m. The horizontal spacing of the
struts was 3.66 m center-to-center.
516 Chapter 10: Braced Cuts
11.28 m
Strut
S1
Fill
7.93 m
S2
17.68 m
S3
Silt
6.09 m
S4
S5
2.44 m Till
1.2 m
Rock
Rock
Figure 10.13 Schematic diagram of test section B for subway extension, MTBA
Because the apparent pressure envelopes available (Section 10.2) are for sand and
clay only, questions may arise about how to treat the fill, silt, and till. Figure 10.14 shows
the apparent pressure envelopes proposed by Peck (1969), considering the soil as sand and
also as clay, to overcome that problem. For the average soil parameters of the profile, the
following values of sa were used to develop the pressure envelopes shown in Figure 10.14.
17.68 m
a =
53.52 kN/m2
(a) Assuming
sand
a = 146.23 kN/m2
(b) Assuming clay
Figure 10.14 Pressure envelopes
(a) assuming sand; (b) assuming clay
10.5 Case Studies of Braced Cuts
517
Sand
sa 5 0.65gHKa
(10.10)
For g 5 17.92 kN>m3, H 5 17.68 m, and Ka 5 0.26,
sa 5 (0.65) (17.92) (17.68) (0.26) 5 53.52 kN>m2
Clay
sa 5 gH c1 2 a
4c
bd
gH
(10.11)
For c 5 42.65 kN>m2,
sa 5 (17.92) (17.68) c1 2
(4) (42.65)
d 5 146.23 kN>m2
(17.92) (17.68)
Table 10.1 shows the variations of the strut load, based on the assumed pressure
envelopes shown in Figure 10.14. Also shown in Table 10.1 are the measured strut loads
in the field and the design strut loads. This comparison indicates that
1. In most cases the measured strut loads differed widely from those predicted. This
result is due primarily to the uncertainties involved in the assumption of the soil
parameters.
2. The actual design strut loads were substantially higher than those measured.
B. Construction of National Plaza (South Half) in Chicago
The construction of the south half of the National Plaza in Chicago required a braced
cut 21.43 m deep. Swatek et al. (1972) reported the case history for this construction.
Figure 10.15 shows a schematic diagram for the braced cut and the subsoil
profile. There were six levels of struts. Table 10.2 gives the actual maximum wale and
strut loads.
Table 10.1 Computed and Measured Strut Loads at Test Section B
Computed load (kip)
Strut
number
Envelope based
on sand
S-1
S-2
S-3
S-4
S-5
810
956
685
480
334
Envelope based
on clay
1023
2580
1868
1299
974
Measured
strut load
(kip)
313
956
1352
1023
1219
518 Chapter 10: Braced Cuts
4.36 m
Existing
Curb wall
Sand fill
30°
17.29 kN/m3
0.305 m
Stiff clay
0.915 m
A
B
Soft silty clay
0
c 19.17 kN/m2
19.97 kN/m3
C
Subway
D
9.76 m
Medium silty clay
0
c 33.54 kN/m2, 20.44 kN/m3
13.11 m
Tough silty clay
0, c 95.83 kN/m2, 21.22 kN/m3
14.94 m
E
F
17.07 m
MZ 38
Sheet piling
–18.90 m
19.51 m
Very tough silty clay
0
c 191.67 kN/m2
21.22 kN/m3
Hardpan
Figure 10.15 Schematic diagram of braced cut—National Plaza of Chicago
Table 10.2 National Plaza Wale and Strut Loads
Strut
level
Elevation
(m)
Load measured
(kN/m)
A
B
C
D
E
F
10.915
21.83
24.57
27.47
210.37
213.57
233.49
386.71
423.20
423.20
423.20
448.0
S2337.8
Figure 10.16 presents a lateral earth-pressure envelope based on the maximum wale
loads measured. To compare the theoretical prediction to the actual observation requires making an approximate calculation. To do so, we convert the clayey soil layers from Elevation
10.305 m ft to 217.07 m to a single equivalent layer in Table 10.3 by using Eq. (10.6).
Now, using Eq. (10.4), we can convert the sand layer located between elevations
14.36 m and 10.305 m and the equivalent clay layer of 17.375 m to one equivalent clay
layer with a thickness of 21.43 m:
1
3gsKsH 2s tan fsr 1 (H 2 Hs )nrqu4
2H
1
d 3 (17.29) (1) (4.055) 2 tan 30 1 (17.375) (0.75) (2 3 51.24) 4
5 c
(2) (21.43)
< 34.99 kN>m2
cav 5
10.5 Case Studies of Braced Cuts
+4.36 m
5.36 m
+0.915 m
A
–1.83 m
B
–4.57 m
C
283.7 kN/m2
–7.47 m
D
Peck's pressure
envelope
–10.37 m
E
16.07 m
Actual pressure envelope
–13.57 m
F
Bottom of cut
–17.07 m
Figure 10.16 Comparison of actual and Peck’s pressure envelopes
Table 10.3 Conversion of Soil Layers using Eq. (10.6)
Elevation
(m)
Thickness,
H (m)
c (kN/m2)
10.305 to 29.67
9.975
19.17
29.67 to 213.11
3.44
33.54
213.11 to 214.94
214.94 to 217.07
1.83
2.13
S17.375
Equivalent
c (kN/m2)
1
3 (9.975) (19.17) 1 (3.44) (33.54)
17.375
1 (1.83) (95.83) (2.13) (191.67)
5 51.24 kN>m2
cav 5
95.83
191.67
Equation (10.7) gives
1
gav 5
3g H 1 g2H2 1 c 1 gnHn )
H 1 1
519
520 Chapter 10: Braced Cuts
1
3(17.29) (4.055) 1 (19.97) (10.065) 1 (20.44) (3.35)
21.43
1 (21.22) (1.83) 1 (21.22) (2.13)4
5 19.77 kN>m3
5
For the equivalent clay layer of 21.43 m,
gavH
(19.77) (21.43)
5
5 12.1 . 4
cav
34.99
Hence the apparent pressure envelope will be of the type shown in Figure 10.6 From Eq. (10.2)
4cav
(4) (34.99)
sa 5 gH c1 2 a
bd 5 (19.77) (21.43) c1 2
d 5 283.7 kN>m2
gavH
(19.77) (12.43)
The pressure envelope is shown in Figure 10.16. The area of this pressure diagram
is 2933 kN/m. Thus Peck’s pressure envelope gives a lateral earth pressure of about
1.8 times that actually observed. This result is not surprising because the pressure envelope provided by Figure 10.6 is an envelope developed considering several cuts made
at different locations. Under actual field conditions, past experience with the behavior of
similar soils can help reduce overdesigning substantially.
10.6
Bottom Heave of a Cut in Clay
Braced cuts in clay may become unstable as a result of heaving of the bottom of the excavation. Terzaghi (1943) analyzed the factor of safety of long braced excavations against bottom
heave. The failure surface for such a case in a homogeneous soil is shown in Figure 10.17.
In the figure, the following notations are used: B 5 width of the cut, H 5 depth of the
cut, T 5 thickness of the clay below the base of excavation, and q 5 uniform surcharge
adjacent to the excavation.
q
e
j
B
c
B
0
c
H
B
g
f
45°
i
45°
T
h
Arc of a
circle
Figure 10.17 Heaving in braced cuts
in clay
10.6 Bottom Heave of a Cut in Clay
521
The ultimate bearing capacity at the base of a soil column with a width of Br can be
given as
qult 5 cNc
(10.12)
where Nc 5 5.7 (for a perfectly rough foundation).
The vertical load per unit area along fi is
q 5 gH 1 q 2
cH
Br
(10.13)
Hence, the factor of safety against bottom heave is
FS 5
qult
5
q
cNc
cH
gH 1 q 2
Br
5
cNc
q
c
¢g 1
2
≤H
H
Br
(10.14)
For excavations of limited length L, the factor of safety can be modified to
cNc ¢1 1 0.2
FS 5
Br
≤
L
q
c
¢g 1
2
≤H
H
Br
(10.15)
where Br 5 T or B>"2 (whichever is smaller).
In 2000, Chang suggested a revision of Eq. (10.15) with the following changes:
1. The shearing resistance along ij may be considered as an increase in resistance
rather than a reduction in loading.
2. In Figure 10.17, fg with a width of Bs at the base of the excavation may be treated
as a negatively loaded footing.
3. The value of the bearing capacity factor Nc should be 5.14 (not 5.7) for a perfectly
smooth footing, because of the restraint-free surface at the base of the excavation.
With the foregoing modifications, Eq. (10.15) takes the form
5.14c ¢1 1
FS 5
0.2Bs
cH
≤ 1
L
Br
gH 1 q
(10.16)
where
Br 5 T if T < B>"2
Br 5 B>"2 if T . B>"2
Bs 5 "2Br
Bjerrum and Eide (1956) compiled a number of case records for the bottom heave
of cuts in clay. Chang (2000) used those records to calculate FS by means of Eq. (10.16);
his findings are summarized in Table 10.4. It can be seen from this table that the actual
field observations agree well with the calculated factors of safety.
522 Chapter 10: Braced Cuts
Table 10.4 Calculated Factors of Safety for Selected Case Records Compiled by Bjerrum and Eide (1956)
and Calculated by Chang (2000)
Site
Pumping station,
Fornebu, Oslo
Storehouse,
Drammen
Sewerage tank,
Drammen
Excavation,
Grey Wedels
Plass, Oslo
Pumping station,
Jernbanetorget,
Oslo
Storehouse, Freia,
Oslo
Subway, Chicago
B
(m)
B,L
H
(m)
H,B
g
(kN , m3)
5.0
1.0
3.0
0.6
17.5
4.8
0
2.4
0.5
19.0
5.5
0.69
3.5
0.64
5.8
0.72
4.5
8.5
0.70
5.0
16
0
0
c
(kN , m2)
q
(kN , m2)
FS
[Eq. (10.16)]
0
1.05
Total failure
12
15
1.05
Total failure
18.0
10
10
0.92
Total failure
0.78
18.0
14
10
1.07
Total failure
6.3
0.74
19.0
22
0
1.26
Partial failure
5.0
11.3
1.00
0.70
19.0
19.0
16
35
0
0
1.10
1.00
Partial failure
Near failure
7.5
Type of
failure
Equation (10.16) is recommended for use in this test. In most cases, a factor of safety
of about 1.5 is recommended.
In homogeneous clay, if FS becomes less than 1.5, the sheet pile is driven deeper.
(See Figure 10.18.) Usually, the depth d is kept less than or equal to B>2, in which case
the force P per unit length of the buried sheet pile (aar and bbr) may be expressed as
(U.S. Department of the Navy, 1971)
P 5 0.7(gHB 2 1.4cH 2 pcB)
for d . 0.47B
(10.17)
B
c
0
H
d
a
b
a
b
P
P
Figure 10.18 Force on the buried length
of sheet pile
10.6 Bottom Heave of a Cut in Clay
523
and
P 5 1.5d ¢gH 2
1.4cH
2 pc ≤
B
for d , 0.47B
(10.18)
Example 10.3
In Figure 10.19. for a braced cut in clay, B 5 3 m, L 5 20 m, H 5 5.5 m, T 5 1.5 m,
g 5 17 kN>m3, c 5 30 kN>m2, and q 5 0. Calculate the factor of safety against heave.
Use Eq. (10.16).
Solution
From Eq. (10.16),
0.2Brr
cH
b1
L
Br
gH 1 q
5.14ca1 1
FS 5
with T 5 1.5 m,
B
!2
So
5
3
!2
T#
Hence, Br 5 T 5 1.5 m, and it follows that
5 2.12 m
B
!2
B rr 5 !2Br 5 ( !2) (1.5) 5 2.12 m
3m
Clay
γ = 17 kN/m3
5.5 m
c = 30 kN/m3
φ=0
1.5 m
Hard stratum
Figure 10.19 Factor of safety against heaving for a braced cut
and
FS 5
(5.14) (30) c1 1
(0.2) (2.12)
(30) (5.5)
d 1
20
1.5
5 2.86
(17) (5.5)
■
524 Chapter 10: Braced Cuts
10.7
Stability of the Bottom of a Cut in Sand
The bottom of a cut in sand is generally stable. When the water table is encountered,
the bottom of the cut is stable as long as the water level inside the excavation is higher
than the groundwater level. In case dewatering is needed (see Figure 10.20), the factor
of safety against piping should be checked. [Piping is another term for failure by
heave, as defined in Section 1.12; see Eq. (1.45).] Piping may occur when a high
hydraulic gradient is created by water flowing into the excavation. To check the factor
of safety, draw flow nets and determine the maximum exit gradient 3imax(exit) 4 that will
occur at points A and B. Figure 10.21 shows such a flow net, for which the maximum
exit gradient is
imax(exit)
h
Nd
h
5
5
a
Nda
(10.19)
where
a 5 length of the flow element at A (or B)
Nd 5 number of drops (Note: in Figure 10.21, Nd 5 8; see also Section 1.11)
The factor of safety against piping may be expressed as
FS 5
icr
imax(exit)
(10.20)
where icr 5 critical hydraulic gradient.
Water
level
Water
level
B
h
L1
A
B
L2
Flow of
water
Impervious layer
L3
Figure 10.20 Stability of the bottom of a cut in sand
10.7 Stability of the Bottom of a Cut in Sand
Water table
525
Water table
h
A
Water
level
B
a
1
8
7
2
6
5 3
4
Impervious layer
Figure 10.21 Determining the factor of safety against piping by drawing a flow net
The relationship for icr was given in Chapter 1 as
icr 5
Gs 2 1
e11
The magnitude of icr varies between 0.9 and 1.1 in most soils, with an average of about 1.
A factor of safety of about 1.5 is desirable.
The maximum exit gradient for sheeted excavations in sands with L3 5 ` can
also be evaluated theoretically (Harr, 1962). (Only the results of these mathematical
derivations will be presented here. For further details, see the original work.) To calculate the maximum exit gradient, examine Figures 10.22 and 10.23 and perform the
following steps:
1. Determine the modulus, m, from Figure 10.22 by obtaining 2L2>B (or B>2L2)
and 2L1>B.
2. With the known modulus and 2L1>B, examine Figure 10.23 and determine
L2iexit(max)>h. Because L2 and h will be known, iexit(max) can be calculated.
3. The factor of safety against piping can be evaluated by using Eq. (10.20).
Marsland (1958) presented the results of model tests conducted to study the influence of seepage on the stability of sheeted excavations in sand. The results were summarized by the U.S. Department of the Navy (1971) in NAVFAC DM-7 and are given in
Figure 10.24a, b, and c. Note that Figure 10.24b is for the case of determining the sheet
pile penetration L2 needed for the required factor of safety against piping when the sand
layer extends to a great depth below the excavation. By contrast, Figure 10.24c represents the case in which an impervious layer lies at a limited depth below the bottom of
the excavation.
526 Chapter 10: Braced Cuts
Text not available due to copyright restrictions
10.7 Stability of the Bottom of a Cut in Sand
0.70
0.65
L2i exit(max)
h
0.60
2L1
B
0.55
=
0
0.5
0.50
1
2
4
8
12
16
20
0.45
0.40
0
0.02
0.04
0.06
0.08
Modulus, m
(a)
0.10
0.12
1.0
1.2
0.6
0.5
L2i exit(max)
h
0.4
2L1
=0
B
0.3
0.5
0.2
1
0.1
0
2
20 12 8 4
16
0.2
0.4
0.6
0.8
Modulus, m
(b)
Figure 10.23 Variation of maximum exit gradient with modulus
(From Groundwater and Seepage, by M. E. Harr. Copyright 1962
by McGraw-Hill. Used with permission.)
527
528 Chapter 10: Braced Cuts
B
Water table
h
Sand
L2
L3
Impervious layer
(a)
2.0
Loose sand
Dense sand
Factor of safety
aganist heave in
loose sand or
piping in dense
sand
2.0
L3 = ∞
1.5
L2
1.0
h
1.5
2.0
1.5
1.0
1.0
0.5
0
0
0.5
1.0
B/2h
(b)
2.0
2.0
1.5
Dense sand of
limited depth:
L3 1.5
L2 1.0
h
L3
=2
h
Factors of
safety agnist
piping
2.0
0.5
2.0
1.5
1.5
1.0
1.0
L3
=1
h
0
0
0.5
1.0
1.5
2.0
B/2h
(c)
Figure 10.24 Influence of seepage on the stability
of sheeted excavation (US Department of Navy, 1971.)
10.8 Lateral Yielding of Sheet Piles and Ground Settlement
529
Example 10.4
In Figure 10.20, let h 5 4.5 m, L1 5 5 m, L2 5 4 m, B 5 5 m, and L3 5 ` . Determine
the factor of safety against piping. Use Figures 10.22 and 10.23.
Solution
We have
2L1
2(5)
5
52
B
5
and
B
5
5
5 0.625
2L2
2(4)
According to Figure 10.22b, for 2L1>B 5 2 and B>2L2 5 0.625,m < 0.033. From
Figure 10.23a, for m 5 0.033 and 2L1>B 5 2, L2iexit(max)>h 5 0.54. Hence,
iexit(max) 5
0.54(h)
5 0.54(4.5)>4 5 0.608
L2
and
FS 5
10.8
icr
1
5
5 1.645
i max (exit)
0.608
■
Lateral Yielding of Sheet Piles
and Ground Settlement
In braced cuts, some lateral movement of sheet pile walls may be expected. (See
Figure 10.25.) The amount of lateral yield (dH ) depends on several factors, the most
important of which is the elapsed time between excavation and the placement of wales
and struts. As discussed before, in several instances the sheet piles (or the soldier piles,
as the case may be) are driven to a certain depth below the bottom of the excavation. The
reason is to reduce the lateral yielding of the walls during the last stages of excavation.
Lateral yielding of the walls will cause the ground surface surrounding the cut to settle.
The degree of lateral yielding, however, depends mostly on the type of soil below the
bottom of the cut. If clay below the cut extends to a great depth and gH>c is less than
about 6, extension of the sheet piles or soldier piles below the bottom of the cut will help
considerably in reducing the lateral yield of the walls.
530 Chapter 10: Braced Cuts
x
Original ground surface
V (max)
Deflected shape
of sheet pile
z
z
H
H
H (max)
Figure 10.25 Lateral yielding of sheet pile and ground settlement
However, under similar circumstances, if gH>c is about 8, the extension of sheet
piles into the clay below the cut does not help greatly. In such circumstances, we may
expect a great degree of wall yielding that could result in the total collapse of the
bracing systems. If a hard layer of soil lies below a clay layer at the bottom of the cut,
the piles should be embedded in the stiffer layer. This action will greatly reduce lateral yield.
The lateral yielding of walls will generally induce ground settlement, dV , around
a braced cut. Such settlement is generally referred to as ground loss. On the basis of several field observations, Peck (1969) provided curves for predicting ground settlement in
various types of soil. (See Figure 10.26.) The magnitude of ground loss varies extensively; however, the figure may be used as a general guide.
Moormann (2004) analyzed about 153 case histories dealing mainly with the excavation in soft clay (that is, undrained shear strength, c < 75 kN>m2). Following is a summary of his analysis relating to dV(max), x r, dH(max), and z r (see Figure 10.25).
•
Maximum Vertical Movement [dV(max)]
dV(max)>H < 0.1 to 10.1% with an average of 1.07% (soft clay)
dV(max)>H < 0 to 0.9% with an average of 0.18% (stiff clay)
dV(max)>H < 0 to 2.43% with an average of 0.33% (non-cohesive soils)
•
Location of dV(max), that is x r (Figure 10.25)
For 70% of all case histories considered, x r < 0.5H.
However, in soft clays, x r may be as much as 2H.
Problems
531
3
A — Sand and soft clay and average
workmanship
B — Very soft to soft clay. Limited in
depth below base of excavation
2
C — Very soft to soft clay. Great depth
below excavation
V
(%)
H
1
C
B
A
0
1
2
3
Distance from the braced wall
H
4
Figure 10.26 Variation of ground settlement with distance (From Peck, R. B. (1969). “Deep
Excavation and Tunneling in Soft Ground,” Proceedings Seventh International Conference on Soil
Mechanics and Foundation Engineering, Mexico City, State-of-the-Art Volume, pp. 225–290. With
permission from ASCE.)
•
Maximum Horizontal Deflection of Sheet Piles, dH(max)
For 40% of excavation in soft clay, 0.5% < dH(max)>H < 1%.
The average value of dH(max)>H is about 0.87%.
In stiff clays, the average value of dH(max)>H is about 0.25%.
In non-cohesive soils, dH(max)>H is about 0.27% of the average.
•
Location of dH(max), that is z r (Figure 10.25)
For deep excavation of soft and stiff cohesive soils, zr>H is about 0.5 to 1.0.
Problems
10.1 Refer to the braced cut shown in Figure P10.1. Given: g 5 16 kN>m3,
fr 5 38°, and c r 5 0. The struts are located at 3.5 m center-to-center in the
plan. Draw the earth-pressure envelope and determine the strut loads at
levels A, B, and C.
532 Chapter 10: Braced Cuts
4.5 m
1m
A
2.5 m
B
Sand
16 kN/m3
38°
c 0
3m
C
1.5 m
Figure P10.1
10.2 For the braced cut described in Problem 10.1, determine the following:
a. The sheet-pile section modulus
b. The section modulus of the wales at level B
Assume that sall 5 170 MN>m2.
10.3 Refer to Fig. P.10.3. Redo Problem 10.1 with g 5 18 kN>m3, fr 5 40°, c r 5 0,
and the center-to-center strut spacing in the plan 5 4 m.
10.4 Determine the sheet-pile section modulus for the braced cut described in
Problem 10.3. Given: sall 5 170 MN>m2.
10.5 Refer to Figure 10.8a. For the braced cut, given H 5 6 m; Hs 5 2.5 m; gs 5
16.5 kN>m3; angle of friction of sand, fsr 5 35°; Hc 5 3.5 m; gc 5 17.5 kN>m3;
and unconfined compression strength of clay layer, qu 5 62 kN>m2.
a. Estimate the average cohesion (cav ) and average unit weight (gav ) for the construction of the earth-pressure envelope.
b. Plot the earth-pressure envelope.
3.5 m
1m
A
2m
B
2m
C
1.5 m
Figure P10.3
Sand
18 kN/m3
38°
c 0
References
533
6m
1m
A
3m
B
c 30 kN/m2
0
17.5 kN/m3
2m
C
1m
Figure P10.7
10.6 Refer to Figure 10.8b, which shows a braced cut in clay. Given: H 5 7.6 m,
H1 5 1.52 m, c1 5 101.8 kN>m2, g1 5 17.45 kN>m3, H2 5 3.04 m,
c2 5 74.56 kN>m2, g2 5 16.83 kN>m3, H3 5 3.04 m, c3 5 80.02 kN>m2, and
g3 5 17.14 kN>m3.
a. Determine the average cohesion (cav ) and average unit weight (gav ) for the
construction of the earth-pressure envelope.
b. Plot the earth-pressure envelope.
10.7 Refer to Figure P10.7. Given: g 5 17.5 kN>m3, c 5 30 kN>m2, and center-tocenter spacing of struts in the plan 5 5 m. Draw the earth-pressure envelope and
determine the strut loads at levels A, B, and C.
10.8 Determine the sheet-pile section modulus for the braced cut described in
Problem 10.7. Use sall 5 170 MN>m.2.
10.9 Redo Problem 10.7 assuming that c 5 60 kN>m2.
10.10 Determine the factor of safety against bottom heave for the braced cut described
in Problem 10.7. Use Eq. (10.16) and assume the length of the cut, L 5 18 m.
10.11 Determine the factor of safety against bottom heave for the braced cut described in
Problem 10.9. Use Eq. (10.15). The length of the cut is 12.5.
References
BJERRUM, L, and EIDE, O. (1956). “Stability of Strutted Excavation in Clay,” Geotechnique, Vol. 6,
No. 1, pp. 32–47.
CHANG, M. F. (2000). “Basal Stability Analysis of Braced Cuts in Clay,” Journal of Geotechnical and
Geoenvironmental Engineering, ASCE, Vol. 126, No. 3, pp. 276–279.
HARR, M. E. (1962). Groundwater and Seepage, McGraw-Hill, New York.
LAMBE, T. W. (1970). “Braced Excavations.” Proceedings of the Specialty Conference on Lateral
Stresses in the Ground and Design of Earth-Retaining Structures, American Society of Civil
Engineers, pp. 149–218.
534 Chapter 10: Braced Cuts
MOORMANN, C. (2004). “Analysis of Wall and Ground Movements Due to Deep Excavations
in Soft Soil Based on New Worldwide Data Base,” Soils and Foundations, Vol. 44, No. 1,
pp. 87–98.
PECK, R. B. (1943). “Earth Pressure Measurements in Open Cuts, Chicago (ILL.) Subway,” Transactions, American Society of Civil Engineers, Vol. 108, pp. 1008–1058.
PECK, R. B. (1969). “Deep Excavation and Tunneling in Soft Ground,” Proceedings Seventh International Conference on Soil Mechanics and Foundation Engineering, Mexico City, State-of-theArt Volume, pp. 225–290.
SWATEK, E. P., JR., ASROW, S. P., and SEITZ, A. (1972). “Performance of Bracing for Deep Chicago
Excavation,” Proceeding of the Specialty Conference on Performance of Earth and Earth Supported Structures, American Society of Civil Engineers, Vol. 1, Part 2, pp. 1303–1322.
TERZAGHI, K. (1943). Theoretical Soil Mechanics, Wiley, New York.
U.S. DEPARTMENT OF THE NAVY (1971). “Design Manual—Soil Mechanics. Foundations, and Earth
Structures.” NAVFAC DM-7, Washington, D.C.
11
11.1
Pile Foundations
Introduction
Piles are structural members that are made of steel, concrete, or timber. They are used to
build pile foundations, which are deep and which cost more than shallow foundations.
(See Chapters 3, 4, and 5.) Despite the cost, the use of piles often is necessary to ensure
structural safety. The following list identifies some of the conditions that require pile
foundations (Vesic, 1977):
1. When one or more upper soil layers are highly compressible and too weak to support the load transmitted by the superstructure, piles are used to transmit the load to
underlying bedrock or a stronger soil layer, as shown in Figure 11.1a. When bedrock
is not encountered at a reasonable depth below the ground surface, piles are used to
transmit the structural load to the soil gradually. The resistance to the applied structural load is derived mainly from the frictional resistance developed at the soil–pile
interface. (See Figure 11.1b.)
2. When subjected to horizontal forces (see Figure 11.1c), pile foundations resist by
bending, while still supporting the vertical load transmitted by the superstructure.
This type of situation is generally encountered in the design and construction of
earth-retaining structures and foundations of tall structures that are subjected to high
wind or to earthquake forces.
3. In many cases, expansive and collapsible soils may be present at the site of a proposed structure. These soils may extend to a great depth below the ground surface.
Expansive soils swell and shrink as their moisture content increases and decreases,
and the pressure of the swelling can be considerable. If shallow foundations are
used in such circumstances, the structure may suffer considerable damage.
However, pile foundations may be considered as an alternative when piles are
extended beyond the active zone, which is where swelling and shrinking occur.
(See Figure 11.1d)
Soils such as loess are collapsible in nature. When the moisture content of these
soils increases, their structures may break down. A sudden decrease in the void ratio of
soil induces large settlements of structures supported by shallow foundations. In such
535
536 Chapter 11: Pile Foundations
Rock
(a)
(b)
(c)
Zone of
erosion
Swelling
soil
Stable
soil
(d)
(e)
(f)
Figure 11.1 Conditions that require the use of pile foundations
cases, pile foundations may be used in which the piles are extended into stable soil
layers beyond the zone where moisture will change.
4. The foundations of some structures, such as transmission towers, offshore
platforms, and basement mats below the water table, are subjected to uplifting
forces. Piles are sometimes used for these foundations to resist the uplifting force.
(See Figure 11.1e.)
5. Bridge abutments and piers are usually constructed over pile foundations to avoid
the loss of bearing capacity that a shallow foundation might suffer because of soil
erosion at the ground surface. (See Figure 11.1f.)
Although numerous investigations, both theoretical and experimental, have been
conducted in the past to predict the behavior and the load-bearing capacity of piles in
granular and cohesive soils, the mechanisms are not yet entirely understood and may
never be. The design and analysis of pile foundations may thus be considered somewhat
of an art as a result of the uncertainties involved in working with some subsoil conditions. This chapter discusses the present state of the art.
11.2 Types of Piles and Their Structural Characteristics
11.2
537
Types of Piles and Their Structural Characteristics
Different types of piles are used in construction work, depending on the type of load to be
carried, the subsoil conditions, and the location of the water table. Piles can be divided into
the following categories: (a) steel piles, (b) concrete piles, (c) wooden (timber) piles, and
(d) composite piles.
Steel Piles
Steel piles generally are either pipe piles or rolled steel H-section piles. Pipe piles can be
driven into the ground with their ends open or closed. Wide-flange and I-section steel
beams can also be used as piles. However, H-section piles are usually preferred because
their web and flange thicknesses are equal. (In wide-flange and I-section beams, the web
thicknesses are smaller than the thicknesses of the flange.) Table 11.1 gives the dimensions of some standard H-section steel piles used in the United States. Table 11.2 shows
selected pipe sections frequency used for piling purposes. In many cases, the pipe piles
are filled with concrete after they have been driven.
The allowable structural capacity for steel piles is
Qall 5 A sfs
(11.1)
where
A s 5 cross-sectional area of the steel
fs 5 allowable stress of steel (<0.33–0.5 f y)
Once the design load for a pile is fixed, one should determine, on the basis of geotechnical considerations, whether Q(design) is within the allowable range as defined by
Eq. 11.1.
When necessary, steel piles are spliced by welding or by riveting. Figure 11.2a
shows a typical splice by welding for an H-pile. A typical splice by welding for a pipe
pile is shown in Figure 11.2b. Figure 11.2c is a diagram of a splice of an H-pile by rivets
or bolts.
When hard driving conditions are expected, such as driving through dense gravel,
shale, or soft rock, steel piles can be fitted with driving points or shoes. Figures 11.2d and
11.2e are diagrams of two types of shoe used for pipe piles.
Steel piles may be subject to corrosion. For example, swamps, peats, and other
organic soils are corrosive. Soils that have a pH greater than 7 are not so corrosive. To offset the effect of corrosion, an additional thickness of steel (over the actual designed
cross-sectional area) is generally recommended. In many circumstances factory-applied
epoxy coatings on piles work satisfactorily against corrosion. These coatings are not easily damaged by pile driving. Concrete encasement of steel piles in most corrosive zones
also protects against corrosion.
Following are some general facts about steel piles:
•
•
Usual length: 15 m to 60 m
Usual load: 300 kN to 1200 kN
538 Chapter 11: Pile Foundations
•
•
Advantages:
a. Easy to handle with respect to cutoff and extension to the desired length
b. Can stand high driving stresses
c. Can penetrate hard layers such as dense gravel and soft rock
d. High load-carrying capacity
Disadvantages:
a. Relatively costly
b. High level of noise during pile driving
c. Subject to corrosion
d. H-piles may be damaged or deflected from the vertical during driving through
hard layers or past major obstructions
Table 11.1 Common H-Pile Sections used in the United States
Moment of
inertia
(m4 3 1026)
Designation,
size (mm) 3
weight (kg/m)
Depth d1
(mm)
Section area
(m2 3 1023)
Flange and web
thickness w
(mm)
Flange width d2
(mm)
lxx
HP 200 3 53
HP 250 3 85
3 62
HP 310 3 125
3 110
3 93
3 79
HP 330 3 149
3 129
3 109
3 89
HP 360 3 174
3 152
3 132
3 108
204
254
246
312
308
303
299
334
329
324
319
361
356
351
346
6.84
10.8
8.0
15.9
14.1
11.9
10.0
19.0
16.5
13.9
11.3
22.2
19.4
16.8
13.8
11.3
14.4
10.6
17.5
15.49
13.1
11.05
19.45
16.9
14.5
11.7
20.45
17.91
15.62
12.82
207
260
256
312
310
308
306
335
333
330
328
378
376
373
371
49.4
123
87.5
271
237
197
164
370
314
263
210
508
437
374
303
y
d2
d1
x
x
w
w
y
lyy
16.8
42
24
89
77.5
63.7
62.9
123
104
86
69
184
158
136
109
11.2 Types of Piles and Their Structural Characteristics
Table 11.2 Selected Pipe Pile Sections
Outside
diameter
(mm)
Wall
thickness
(mm)
219
3.17
21.5
4.78
32.1
5.56
37.3
7.92
52.7
4.78
37.5
5.56
43.6
6.35
49.4
4.78
44.9
5.56
52.3
6.35
59.7
4.78
60.3
5.56
70.1
6.35
79.8
5.56
80
6.35
90
7.92
112
5.56
88
6.35
100
7.92
125
6.35
121
7.92
150
9.53
179
12.70
238
254
305
406
457
508
610
Area of
steel
(cm2)
539
540 Chapter 11: Pile Foundations
Weld
Weld
(b)
(a)
(c)
Weld
Weld
(d)
(e)
Figure 11.2 Steel piles: (a) splicing of H-pile by welding; (b) splicing of pipe pile by welding;
(c) splicing of H-pile by rivets and bolts; (d) flat driving point of pipe pile; (e) conical driving
point of pipe pile
Concrete Piles
Concrete piles may be divided into two basic categories: (a) precast piles and (b) cast-in-situ
piles. Precast piles can be prepared by using ordinary reinforcement, and they can be square
or octagonal in cross section. (See Figure 11.3.) Reinforcement is provided to enable the
pile to resist the bending moment developed during pickup and transportation, the vertical load, and the bending moment caused by a lateral load. The piles are cast to desired
lengths and cured before being transported to the work sites.
Some general facts about concrete piles are as follows:
•
•
•
Usual length: 10 m to 15 m
Usual load: 300 kN to 3000 kN
Advantages:
a. Can be subjected to hard driving
b. Corrosion resistant
c. Can be easily combined with a concrete superstructure
11.2 Types of Piles and Their Structural Characteristics
541
2D
Square pile
Octagonal pile
D
D
Figure 11.3 Precast piles with ordinary
reinforcement
•
Disadvantages:
a. Difficult to achieve proper cutoff
b. Difficult to transport
Precast piles can also be prestressed by the use of high-strength steel pre-stressing
cables. The ultimate strength of these cables is about 1800 MN>m2. During casting of the
piles, the cables are pretensioned to about 900 to 1300 MN>m2, and concrete is poured
around them. After curing, the cables are cut, producing a compressive force on the pile
section. Table 11.3 gives additional information about prestressed concrete piles with
square and octagonal cross sections.
Some general facts about precast prestressed piles are as follows:
•
•
•
Usual length: 10 m to 45 m
Maximum length: 60 m
Maximum load: 7500 kN to 8500 kN
The advantages and disadvantages are the same as those of precast piles.
Cast-in-situ, or cast-in-place, piles are built by making a hole in the ground and then
filling it with concrete. Various types of cast-in-place concrete piles are currently used in
construction, and most of them have been patented by their manufacturers. These piles may
be divided into two broad categories: (a) cased and (b) uncased. Both types may have a
pedestal at the bottom.
Cased piles are made by driving a steel casing into the ground with the help of a
mandrel placed inside the casing. When the pile reaches the proper depth the mandrel is
withdrawn and the casing is filled with concrete. Figures 11.4a, 11.4b, 11.4c, and 11.4d
show some examples of cased piles without a pedestal. Figure 11.4e shows a cased pile
with a pedestal. The pedestal is an expanded concrete bulb that is formed by dropping a
hammer on fresh concrete.
Some general facts about cased cast-in-place piles are as follows:
•
•
•
•
Usual length: 5 m to 15 m
Maximum length: 30 m to 40 m
Usual load: 200 kN to 500 kN
Approximate maximum load: 800 kN
542 Chapter 11: Pile Foundations
Table 11.3 Typical Prestressed Concrete Pile in Use
Design bearing
capacity (kN)
Pile
shapea
D
(mm)
Area of cross
section
(cm2)
Perimeter
(mm)
12.7-mm
diameter
11.1-mm
diameter
Minimum
effective
prestress
force (kN)
S
O
S
O
S
O
S
O
S
O
S
O
S
O
S
O
254
254
305
305
356
356
406
406
457
457
508
508
559
559
610
610
645
536
929
768
1265
1045
1652
1368
2090
1729
2581
2136
3123
2587
3658
3078
1016
838
1219
1016
1422
1168
1626
1346
1829
1524
2032
1677
2235
1854
2438
2032
4
4
5
4
6
5
8
7
10
8
12
10
15
12
18
15
4
4
6
5
8
7
11
9
13
11
16
14
20
16
23
19
312
258
449
369
610
503
796
658
1010
836
1245
1032
1508
1250
1793
1486
a
Number of strands
Strength
of concrete
(MN/m2)
Section
modulus
(m3 3 10 23)
34.5
41.4
2.737
1.786
4.719
3.097
7.489
4.916
11.192
7.341
15.928
10.455
21.844
14.355
29.087
19.107
37.756
34.794
556
462
801
662
1091
901
1425
1180
1803
1491
2226
1842
2694
2231
3155
2655
778
555
962
795
1310
1082
1710
1416
2163
1790
2672
2239
3232
2678
3786
3186
S 5 square section; O 5 octagonal section
Wire
spiral
Prestressed
strand
D
D
Wire
spiral
•
•
•
Prestressed
strand
Advantages:
a. Relatively cheap
b. Allow for inspection before pouring concrete
c. Easy to extend
Disadvantages:
a. Difficult to splice after concreting
b. Thin casings may be damaged during driving
Allowable load:
Qall 5 A sfs 1 A cfc
where
A s 5 area of cross section of steel
A c 5 area of cross section of concrete
fs 5 allowable stress of steel
fc 5 allowable stress of concrete
(11.2)
11.2 Types of Piles and Their Structural Characteristics
Monotube or
Union Metal Pile
Western
Cased Pile
Thin, fluted, tapered
steel casing driven
without mandrel
Thin metal casing
Raymond
Step-Taper Pile
Corrugated thin
cylindrical casing
Maximum usual
length: 30 m
(a)
Maximum usual
length: 30 m –40 m
Maximum usual
length: 40 m
(b)
(c)
Seamless Pile or
Armco Pile
Franki Cased
Pedestal Pile
Thin metal casing
Straight steel pile
casing
Maximum usual
length: 30 m– 40 m
(d)
Maximum usual
length: 30 m– 40 m
(e)
Western Uncased
Pile without
Pedestal
Franki Uncased
Pedestal Pile
Maximum usual
length: 30 m –40 m
Maximum usual
length: 15 m –20 m
(f)
Figure 11.4 Cast-in-place concrete piles
543
(g)
544 Chapter 11: Pile Foundations
Figures 11.4f and 11.4g are two types of uncased pile, one with a pedestal and the
other without. The uncased piles are made by first driving the casing to the desired depth and
then filling it with fresh concrete. The casing is then gradually withdrawn.
Following are some general facts about uncased cast-in-place concrete piles:
•
•
•
•
•
•
•
Usual length: 5 m to 15 m
Maximum length: 30 m to 40 m
Usual load: 300 kN to 500 kN
Approximate maximum load: 700 kN
Advantages:
a. Initially economical
b. Can be finished at any elevation
Disadvantages:
a. Voids may be created if concrete is placed rapidly
b. Difficult to splice after concreting
c. In soft soils, the sides of the hole may cave in, squeezing the concrete
Allowable load:
Qall 5 A cfc
(11.3)
where
A c 5 area of cross section of concrete
fc 5 allowable stress of concrete
Timber Piles
Timber piles are tree trunks that have had their branches and bark carefully trimmed off.
The maximum length of most timber piles is 10 to 20 m. To qualify for use as a pile, the
timber should be straight, sound, and without any defects. The American Society of Civil
Engineers’ Manual of Practice, No. 17 (1959), divided timber piles into three classes:
1. Class A piles carry heavy loads. The minimum diameter of the butt should be 356 mm.
2. Class B piles are used to carry medium loads. The minimum butt diameter should be
305 to 330 mm.
3. Class C piles are used in temporary construction work. They can be used permanently for structures when the entire pile is below the water table. The minimum
butt diameter should be 305 mm.
In any case, a pile tip should not have a diameter less than 150 mm.
Timber piles cannot withstand hard driving stress; therefore, the pile capacity is generally limited. Steel shoes may be used to avoid damage at the pile tip (bottom). The tops
of timber piles may also be damaged during the driving operation. The crushing of the
wooden fibers caused by the impact of the hammer is referred to as brooming. To avoid
damage to the top of the pile, a metal band or a cap may be used.
Splicing of timber piles should be avoided, particularly when they are expected to
carry a tensile load or a lateral load. However, if splicing is necessary, it can be done by
using pipe sleeves (see Figure 11.5a) or metal straps and bolts (see Figure 11.5b). The
length of the sleeve should be at least five times the diameter of the pile. The butting ends
should be cut square so that full contact can be maintained. The spliced portions should be
carefully trimmed so that they fit tightly to the inside of the pipe sleeve. In the case of
metal straps and bolts, the butting ends should also be cut square. The sides of the spliced
portion should be trimmed plane for putting the straps on.
11.2 Types of Piles and Their Structural Characteristics
545
Metal
strap
Metal
sleeve
Ends cut
square
Ends
cut
square
Metal
strap
(a)
(b)
Figure 11.5 Splicing of
timber piles: (a) use of pipe
sleeves; (b) use of metal
straps and bolts
Timber piles can stay undamaged indefinitely if they are surrounded by saturated
soil. However, in a marine environment, timber piles are subject to attack by various organisms and can be damaged extensively in a few months. When located above the water
table, the piles are subject to attack by insects. The life of the piles may be increased by
treating them with preservatives such as creosote.
The allowable load-carrying capacity of wooden piles is
Qall 5 A pfw
(11.4)
where
A p 5 average area of cross section of the pile
fw 5 allowable stress on the timber
The following allowable stresses are for pressure-treated round timber piles made from
Pacific Coast Douglas fir and Southern pine used in hydraulic structures (ASCE, 1993):
Pacific Coast Douglas Fir
•
•
•
•
Compression parallel to grain: 6.04 MN>m2
Bending: 11.7 MN>m2
Horizontal shear: 0.66 MN>m2
Compression perpendicular to grain: 1.31 MN>m2
Southern Pine
• Compression parallel to grain: 5.7 MN>m2
• Bending: 11.4 MN>m2
546 Chapter 11: Pile Foundations
• Horizontal shear: 0.62 MN>m2
• Compression perpendicular to grain: 1.41 MN>m2
The usual length of wooden piles is 5 m to 15 m. The maximum length is about 30 m
to 40 m (100 ft to 130 ft). The usual load carried by wooden piles is 300 kN to 500 kN.
Composite Piles
The upper and lower portions of composite piles are made of different materials. For
example, composite piles may be made of steel and concrete or timber and concrete.
Steel-and-concrete piles consist of a lower portion of steel and an upper portion of castin-place concrete. This type of pile is used when the length of the pile required for
adequate bearing exceeds the capacity of simple cast-in-place concrete piles. Timber-andconcrete piles usually consist of a lower portion of timber pile below the permanent
water table and an upper portion of concrete. In any case, forming proper joints between
two dissimilar materials is difficult, and for that reason, composite piles are not
widely used.
11.3
Estimating Pile Length
Selecting the type of pile to be used and estimating its necessary length are fairly difficult
tasks that require good judgment. In addition to being broken down into the classification
given in Section 11.2, piles can be divided into three major categories, depending on their
lengths and the mechanisms of load transfer to the soil: (a) point bearing piles, (b) friction
piles, and (c) compaction piles.
Point Bearing Piles
If soil-boring records establish the presence of bedrock or rocklike material at a site within a
reasonable depth, piles can be extended to the rock surface. (See Figure 11.6a.) In this case,
the ultimate capacity of the piles depends entirely on the load-bearing capacity of the underlying material; thus, the piles are called point bearing piles. In most of these cases, the necessary length of the pile can be fairly well established.
If, instead of bedrock, a fairly compact and hard stratum of soil is encountered at a
reasonable depth, piles can be extended a few meters into the hard stratum. (See Figure
11.6b.) Piles with pedestals can be constructed on the bed of the hard stratum, and the ultimate pile load may be expressed as
Qu 5 Qp 1 Qs
(11.5)
where
Qp 5 load carried at the pile point
Qs 5 load carried by skin friction developed at the side of the pile (caused by shearing
resistance between the soil and the pile)
If Qs is very small,
Qs < Qp
(11.6)
In this case, the required pile length may be estimated accurately if proper subsoil exploration records are available.
11.3 Estimating Pile Length
Qu
Qu
Weak
soil
L
Qu
Qs
L
Qs
Weak
soil
Lb
Qp Rock
Qu Qp
(a)
547
L
Weak
soil
Strong
soil
layer
Qp
Qu Q p
Qp
Qu Qs
Lb depth of penetration
into bearing stratum
(b)
(c)
Figure 11.6 (a) and (b) Point bearing piles; (c) friction piles
Friction Piles
When no layer of rock or rocklike material is present at a reasonable depth at a site, point
bearing piles become very long and uneconomical. In this type of subsoil, piles are driven through the softer material to specified depths. (See Figure 11.6c.) The ultimate load
of the piles may be expressed by Eq. (11.5). However, if the value of Qp is relatively
small, then
Qu < Qs
(11.7)
These piles are called friction piles, because most of their resistance is derived
from skin friction. However, the term friction pile, although used often in the
literature, is a misnomer: In clayey soils, the resistance to applied load is also
caused by adhesion.
The lengths of friction piles depend on the shear strength of the soil, the applied
load, and the pile size. To determine the necessary lengths of these piles, an engineer
needs a good understanding of soil–pile interaction, good judgment, and experience.
Theoretical procedures for calculating the load-bearing capacity of piles are presented
later in the chapter.
Compaction Piles
Under certain circumstances, piles are driven in granular soils to achieve proper compaction of soil close to the ground surface. These piles are called compaction piles. The
lengths of compaction piles depend on factors such as (a) the relative density of the soil
before compaction, (b) the desired relative density of the soil after compaction, and
(c) the required depth of compaction. These piles are generally short; however, some
field tests are necessary to determine a reasonable length.
548 Chapter 11: Pile Foundations
11.4
Installation of Piles
Most piles are driven into the ground by means of hammers or vibratory drivers. In special
circumstances, piles can also be inserted by jetting or partial augering. The types of hammer
used for pile driving include (a) the drop hammer, (b) the single-acting air or steam hammer,
(c) the double-acting and differential air or steam hammer, and (d) the diesel hammer. In the
driving operation, a cap is attached to the top of the pile. A cushion may be used between the
pile and the cap. The cushion has the effect of reducing the impact force and spreading it over
a longer time; however, the use of the cushion is optional. A hammer cushion is placed on
the pile cap. The hammer drops on the cushion.
Figure 11.7 illustrates various hammers. A drop hammer (see Figure 11.7a) is raised
by a winch and allowed to drop from a certain height H. It is the oldest type of hammer
used for pile driving. The main disadvantage of the drop hammer is its slow rate of blows.
The principle of the single-acting air or steam hammer is shown in Figure 11.7b. The striking part, or ram, is raised by air or steam pressure and then drops by gravity. Figure 11.7c
shows the operation of the double-acting and differential air or steam hammer. Air or steam
is used both to raise the ram and to push it downward, thereby increasing the impact velocity of the ram. The diesel hammer (see Figure 11.7d) consists essentially of a ram, an anvil
block, and a fuel-injection system. First the ram is raised and fuel is injected near the anvil.
Then the ram is released. When the ram drops, it compresses the air–fuel mixture, which
ignites. This action, in effect, pushes the pile downward and raises the ram. Diesel hammers work well under hard driving conditions. In soft soils, the downward movement of
the pile is rather large, and the upward movement of the ram is small. This differential may
Exhaust
Cylinder
Intake
Ram
Ram
Hammer
cushion
Hammer
cushion
Pile
cap
Pile
cap
Pile
cushion
Pile
cushion
Pile
Pile
(a)
(b)
Figure 11.7 Pile-driving equipment: (a) drop hammer; (b) single-acting air or steam hammer
11.4 Installation of Piles
549
Exhaust
and
Intake
Cylinder
Exhaust
and
Intake
Ram
Ram
Anvil
Hammer
cushion
Pile cushion
Pile
cap
Pile cap
Pile
cushion
Hammer
cushion
Pile
Pile
(d)
(c)
Static
weight
Oscillator
Clamp
Pile
(e)
(f)
Figure 11.7 (continued) Pile-driving equipment: (c) double-acting and differential air or steam
hammer; (d) diesel hammer; (e) vibratory pile driver; (f) photograph of a vibratory pile driver
(Courtesy of Michael W. O’Neill, University of Houston)
550 Chapter 11: Pile Foundations
Table 11.4 Examples of Commercially Available Pile-Driving Hammers
Maker of
hammer†
Model
No.
Rated energy
kN-m
Blows/min
Ram weight
kN
V
M
M
M
R
R
400C
S-20
S-8
S-5
5/O
2/O
Single acting
153.9
81.3
35.3
22.0
77.1
44.1
100
60
53
60
44
50
177.9
89.0
35.6
22.2
77.8
44.5
V
V
V
V
R
200C
140C
80C
65C
150C
Double acting
or
differential
68.1
48.8
33.1
26.0
66.1
98
103
111
117
95–105
89.0
62.3
35.6
28.9
66.7
V
V
M
M
4N100
IN100
DE40
DE30
Diesel
58.8
33.4
43.4
30.4
50–60
50–60
48
48
23.5
13.3
17.8
12.5
Hammer type
†
V—Vulcan Iron Works, Florida
M—McKiernan-Terry, New Jersey
R—Raymond International, Inc., Texas
not be sufficient to ignite the air–fuel system, so the ram may have to be lifted manually.
Table 11.4 provides some examples of commercially available pile-driving hammers.
The principles of operation of a vibratory pile driver are shown in Figure 11.7e. This
driver consists essentially of two counterrotating weights. The horizontal components of the
centrifugal force generated as a result of rotating masses cancel each other. As a result, a
sinusoidal dynamic vertical force is produced on the pile and helps drive the pile downward.
Figure 11.7f is a photograph of a vibratory pile driver. Figure 11.8 shows a pile-driving
operation in the field.
Jetting is a technique that is sometimes used in pile driving when the pile needs to
penetrate a thin layer of hard soil (such as sand and gravel) overlying a layer of softer soil.
In this technique, water is discharged at the pile point by means of a pipe 50 to 75 mm
in diameter to wash and loosen the sand and gravel.
Piles driven at an angle to the vertical, typically 14 to 20°, are referred to as batter
piles. Batter piles are used in group piles when higher lateral load-bearing capacity is
required. Piles also may be advanced by partial augering, with power augers (see Chapter 2)
used to predrill holes part of the way. The piles can then be inserted into the holes and driven to the desired depth.
Piles may be divided into two categories based on the nature of their placement:
displacement piles and nondisplacement piles. Driven piles are displacement piles,
because they move some soil laterally; hence, there is a tendency for densification of soil
surrounding them. Concrete piles and closed-ended pipe piles are high-displacement
piles. However, steel H-piles displace less soil laterally during driving, so they are lowdisplacement piles. In contrast, bored piles are nondisplacement piles because their placement causes very little change in the state of stress in the soil.
11.5 Load Transfer Mechanism
551
Figure 11.8 A pile-driving operation in the field (Courtesy of E. C. Shin, University
of Incheon, Korea)
11.5
Load Transfer Mechanism
The load transfer mechanism from a pile to the soil is complicated. To understand it, consider a pile of length L, as shown in Figure 11.9a. The load on the pile is gradually
increased from zero to Q(z50) at the ground surface. Part of this load will be resisted by
552 Chapter 11: Pile Foundations
Qu
Q(z 0)
Q(z 0)
Q
z
z
Q(z)
Q1
L
Q(z)
1
2
Q2
Q2
Qp
Qs
(a)
(b)
Unit
frictional
resistance
z0
f(z) Qu
Q(z)
p z
Qs
L
Pile tip
Zone
II
Zone I
Zone
II
zL
Qp
(c)
(d)
(e)
Figure 11.9 Load transfer mechanism for piles
the side friction developed along the shaft, Q1 , and part by the soil below the tip of the pile,
Q2 . Now, how are Q1 and Q2 related to the total load? If measurements are made to obtain
the load carried by the pile shaft, Q(z) , at any depth z, the nature of the variation found will
be like that shown in curve 1 of Figure 11.9b. The frictional resistance per unit area at any
depth z may be determined as
f(z) 5
DQ(z)
(p) ( Dz)
(11.8)
11.5 Load Transfer Mechanism
553
where p 5 perimeter of the cross section of the pile. Figure 11.9c shows the variation of
f(z) with depth.
If the load Q at the ground surface is gradually increased, maximum frictional resistance along the pile shaft will be fully mobilized when the relative displacement between
the soil and the pile is about 5 to 10 mm, irrespective of the pile size and length L. However,
the maximum point resistance Q2 5 Qp will not be mobilized until the tip of the pile has
moved about 10 to 25% of the pile width (or diameter). (The lower limit applies to driven
piles and the upper limit to bored piles). At ultimate load (Figure 11.9d and curve 2 in
Figure 11.9b), Q(z50) 5 Qu . Thus,
Q1 5 Qs
and
Q2 5 Qp
The preceding explanation indicates that Qs (or the unit skin friction, f, along the pile
shaft) is developed at a much smaller pile displacement compared with the point
resistance, Qp . In order to demonstrate this point, let us consider the results of a pile load
test conducted in the field by Mansur and Hunter (1970). The details of the pile and subsoil
conditions are as follow:
Type of pile: Steel pile with 406 mm outside diameter with 8.15 mm
wall thickness
Type of subsoil: Sand
Length of pile embedment: 16.8 m
Figure 11.10a shows the load test results, which is a plot of load at the top of the pile
3Q(z50) 4 versus settlement(s). Figure 11.10b shows the plot of the load carried by the pile
shaft 3Q(z) 4 at any depth. It was reported by Mansur and Hunter (1970) that, for this test,
at failure
Qu < 1601 kN
Qp < 416 kN
and
Qs < 1185 kN
Now, let us consider the load distribution in Figure 11.10b when the pile settlement(s) is
about 2.5 mm. For this condition,
Q(z50) < 667 kN
Q2
< 93 kN
Q1
< 574 kN
Hence, at s 5 2.5 mm,
Q2
93
5
(100) 5 22.4%
Qp
416
and
Q1
574
5
(100) 5 48.4%
Qs
1185
Thus, it is obvious that the skin friction is mobilized faster at low settlement levels
as compared to the point load.
554 Chapter 11: Pile Foundations
Load at the top of pile, Q(z = 0) (kN)
0
400
800
1200
1600
Q(z = 0) (kN)
2000 2200
0
400
800
1200
1600
2000
0
5
3
s = 2.5 mm
s = 5 mm
6
Depth, z (m)
Settlement, s (mm)
s = 11 mm
10
15
9
20
12
25
15
30
16.8
35
(a)
(b)
Figure 11.10 Load test results on a pipe pile in sand (Based on Mansur and Hunter, 1970)
At ultimate load, the failure surface in the soil at the pile tip (a bearing capacity failure caused by Qp) is like that shown in Figure 11.9e. Note that pile foundations are deep
foundations and that the soil fails mostly in a punching mode, as illustrated previously in
Figures 3.1c and 3.3. That is, a triangular zone, I, is developed at the pile tip, which is pushed
downward without producing any other visible slip surface. In dense sands and stiff clayey
soils, a radial shear zone, II, may partially develop. Hence, the load displacement curves of
piles will resemble those shown in Figure 3.1c.
11.6
Equations for Estimating Pile Capacity
The ultimate load-carrying capacity Qu of a pile is given by the equation
Qu 5 Qp 1 Qs
(11.9)
where
Qp 5 load-carrying capacity of the pile point
Qs 5 frictional resistance (skin friction) derived from the soil–pile interface (see
Figure 11.11)
Numerous published studies cover the determination of the values of Qp and Qs . Excellent
reviews of many of these investigations have been provided by Vesic (1977), Meyerhof (1976),
and Coyle and Castello (1981). These studies afford an insight into the problem of determining
the ultimate pile capacity.
11.6 Equations for Estimating Pile Capacity
555
Qu
Steel
Soil plug
Qs
D
(b) Open-Ended Pipe Pile Section
L Lb
D
Steel
q
Soil plug
d1
Qp
L length of embedment
Lb length of embedment in
bearing stratum
(a)
d2
(c) H-Pile Section
(Note: Ap area of steel soil plug)
Figure 11.11 Ultimate load-carrying capacity of pile
Point Bearing Capacity, Qp
The ultimate bearing capacity of shallow foundations was discussed in Chapter 3.
According to Terzaghi’s equations,
qu 5 1.3crNc 1 qNq 1 0.4gBNg
(for shallow square foundations)
qu 5 1.3crNc 1 qNq 1 0.3gBNg
(for shallow circular foundations)
and
Similarly, the general bearing capacity equation for shallow foundations was given in
Chapter 3 (for vertical loading) as
qu 5 crNcFcsFcd 1 qNqFqsFqd 1 12gBNgFgsFgd
Hence, in general, the ultimate load-bearing capacity may be expressed as
qu 5 crN *c 1 qN *q 1 gBN *g
(11.10)
where N c* , N q* , and N g* are the bearing capacity factors that include the necessary shape
and depth factors.
Pile foundations are deep. However, the ultimate resistance per unit area developed at
the pile tip, qp , may be expressed by an equation similar in form to Eq. (11.10), although the
values of N c* , N q* , and N g* will change. The notation used in this chapter for the width of a
pile is D. Hence, substituting D for B in Eq. (11.10) gives
qu 5 qp 5 crN c* 1 qN q* 1 gDN g*
(11.11)
556 Chapter 11: Pile Foundations
Because the width D of a pile is relatively small, the term gDN g* may be dropped from
the right side of the preceding equation without introducing a serious error; thus,
we have
qp 5 crN *c 1 qrN *q
(11.12)
Note that the term q has been replaced by qr in Eq. (11.12), to signify effective vertical
stress. Thus, the point bearing of piles is
Qp 5 Ap qp 5 Ap (crN *c 1 qrN *q )
(11.13)
where
A p 5 area of pile tip
cr 5 cohesion of the soil supporting the pile tip
qp 5 unit point resistance
qr 5 effective vertical stress at the level of the pile tip
N c* , N q* 5 the bearing capacity factors
Frictional Resistance, Qs
The frictional, or skin, resistance of a pile may be written as
Qs 5 S p DLf
(11.14)
where
p 5 perimeter of the pile section
DL 5 incremental pile length over which p and f are taken to be constant
f 5 unit friction resistance at any depth z
The various methods for estimating Qp and Qs are discussed in the next several sections. It needs to be reemphasized that, in the field, for full mobilization of the point resistance (Qp ), the pile tip must go through a displacement of 10 to 25% of the pile
width (or diameter).
Allowable Load, Qall
After the total ultimate load-carrying capacity of a pile has been determined by summing
the point bearing capacity and the frictional (or skin) resistance, a reasonable factor of
safety should be used to obtain the total allowable load for each pile, or
Qall 5
Qu
FS
where
Qall 5 allowable load-carrying capacity for each pile
FS 5 factor of safety
The factor of safety generally used ranges from 2.5 to 4, depending on the uncertainties
surrounding the calculation of ultimate load.
11.7 Meyerhof’s Method for Estimating Qp
11.7
557
Meyerhof’s Method for Estimating Qp
Sand
The point bearing capacity, qp , of a pile in sand generally increases with the depth of embedment in the bearing stratum and reaches a maximum value at an embedment ratio of
Lb>D 5 (Lb>D) cr . Note that in a homogeneous soil Lb is equal to the actual embedment
length of the pile, L. However, where a pile has penetrated into a bearing stratum, Lb , L.
Beyond the critical embedment ratio, (Lb>D) cr , the value of qp remains constant (qp 5 ql ).
That is, as shown in Figure 11.12 for the case of a homogeneous soil, L 5 Lb .
For piles in sand, cr 5 0, and Eq. (11.13) simpifies to
Qp 5 Ap qp 5 Ap qrN *q
(11.15)
The variation of N *q with soil friction angle fr is shown in Figure 11.13. The interpolated
values of N *q for various friction angles are also given in Table 11.5. However, Qp should
not exceed the limiting value A pql ; that is,
Qp 5 A pqrN *q < A pql
Unit point
resistance,
qp
1000
800
600
400
200
100
80
60
N*q
Figure 11.13 Variation of the maximum values
of N *q with soil friction angle fr (From
Meyerhof, G. G. (1976). “Bearing Capacity and
Settlement of Pile Foundations,” Journal of the
Geotechnical Engineering Division, American
Society of Civil Engineers, Vol. 102, No. GT3,
pp. 197–228. With permission from ASCE.)
(11.16)
40
20
(Lb /D)cr
N*q
10
8
6
4
qp ql
L/D Lb /D
Figure 11.12 Nature of variation
of unit point resistance in a
homogeneous sand
2
1
0
10
20
30
Soil friction angle, (deg)
40
45
558 Chapter 11: Pile Foundations
Table 11.5 Interpolated Values of
N *q Based on Meyerhof’s Theory
Soil friction
angle, f (deg)
Nq*
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
12.4
13.8
15.5
17.9
21.4
26.0
29.5
34.0
39.7
46.5
56.7
68.2
81.0
96.0
115.0
143.0
168.0
194.0
231.0
276.0
346.0
420.0
525.0
650.0
780.0
930.0
The limiting point resistance is
ql 5 0.5 paN *q tan fr
(11.17)
where
pa 5 atmospheric pressure (5100 kN>m2)
fr 5 effective soil friction angle of the bearing stratum
A good example of the concept of the critical embedment ratio can be found from the
field load tests on a pile in sand at the Ogeechee River site reported by Vesic (1970). The
pile tested was a steel pile with a diameter of 457 mm. Table 11.6 shows the ultimate resistance at various depths. Figure 11.14 shows the plot of qp with depth obtained from the field
tests along with the range of standard penetration resistance at the site. From the figure, the
following observations can be made.
1. There is a limiting value of qp. For the tests under consideration, it is about 12,000 kN/m2.
2. The (L>D)cr value is about 16 to 18.
11.7 Meyerhof’s Method for Estimating Qp
559
Table 11.6 Ultimate Point Resistance, qp, of Test Pile at the Ogeechee
River Site As reported by Vesic (1970)
Pile diameter,
D (m)
Depth of embedment,
L (m)
L ,D
qp
(kN , m2)
0.457
0.457
0.457
0.457
0.457
3.02
6.12
8.87
12.0
15.00
6.61
13.39
19.4
26.26
32.82
3,304
9,365
11,472
11,587
13,971
0
Pile point resistance, qp(kN/m2)
4000
8000
12,000 16,000
20,000
0
2
Pile point
resistance
4
Depth (m)
6
Range of
N60 at site
8
10
12
14
0
10
20
30
N60
40
50
Figure 11.14 Vesic’s pile test
(1970) result—variation of qp and
N60 with depth
3. The average N60 value is about 30 for L>D > (L>D) cr. Using Eq. (11.37), the
limiting point resistance is 4pa N60 5 (4)(100)(30) 5 12,000 kN/m2. This value is
generally consistent with the field observation.
Clay (f 5 0)
For piles in saturated clays under undrained conditions (f 5 0), the net ultimate load can
be given as
Qp < N *c cuA p 5 9cuA p
where cu 5 undrained cohesion of the soil below the tip of the pile.
(11.18)
560 Chapter 11: Pile Foundations
11.8
Vesic’s Method for Estimating Q p
Sand
Vesic (1977) proposed a method for estimating the pile point bearing capacity based on the
theory of expansion of cavities. According to this theory, on the basis of effective stress
parameters, we may write
Qp 5 A pqp 5 A p sor N *s
(11.19)
where
sor 5 mean effective normal ground stress at the level of the pile point
1 1 2Ko
≤qr
3
Ko 5 earth pressure coefficient at rest 5 1 2 sin fr
5¢
(11.20)
(11.21)
and
N *s 5 bearing capacity factor
Note that Eq. (11.19) is a modification of Eq. (11.15) with
N s* 5
3N *q
(1 1 2Ko )
(11.22)
According to Vesic’s theory,
N s* 5 f(Irr )
(11.23)
where Irr 5 reduced rigidity index for the soil. However,
Irr 5
Ir
1 1 Ir D
(11.24)
where
Ir 5 rigidity index 5
Es
Gs
5
2(1 1 ms ) qr tan fr
qr tan fr
(11.25)
Es 5 modulus of elasticity of soil
ms 5 Poisson’s ratio of soil
Gs 5 shear modulus of soil
D 5 average volumatic strain in the plastic zone below the pile point
The general ranges of Ir for various soils are
Sand(relative density 5 50% to 80%): 75 to 150
Silt : 50 to 75
In order to estimate Ir [Eq. (11.25)] and hence Irr [Eq. (11.24)], the following approximations may be used (Chen and Kulhawy, 1994)
11.8 Vesic’s Method for Estimating Qp
Es
5m
pa
561
(11.26)
where
pa 5 atmospheric pressure ( < 100 kN>m2 )
100 to 200(loose soil)
m 5 c200 to 500(medium dense soil)
500 to 1000(dense soil)
ms 5 0.1 1 0.3a
fr 2 25
b (for 25° # f r # 45°)
20
(11.27)
fr 2 25 qr
b
pa
20
(11.28)
D 5 0.005a1 2
On the basis of cone penetration tests in the field, Baldi et al. (1981) gave the following correlations for Ir :
Ir 5
and
300
Fr (%)
Ir 5
170
Fr (%)
(for mechanical cone penetration)
(11.29)
(for electric cone penetration)
(11.30)
For the definition of Fr , see Eq. (2.41). Table 11.7 gives the values of N s* for various
values of Irr and fr.
Clay (f 5 0)
In saturated clay (f 5 0 condition), the net ultimate point bearing capacity of a pile can
be approximated as
Qp 5 A pqp 5 A pcuN c*
(11.31)
where cu 5 undrained cohesion
According to the expansion of cavity theory of Vesic (1977),
N *c 5
4
p
(ln Irr 1 1) 1 1 1
3
2
(11.32)
The variations of N c* with Irr for f 5 0 condition are given in Table 11.8.
Now, referring to Eq. (11.24) for saturated clay with no volume change, D 5 0.
Hence,
Irr 5 Ir
(11.33)
562
12.12
13.18
14.33
15.57
16.90
18.24
19.88
21.55
23.34
25.28
27.36
29.60
32.02
34.63
37.44
40.47
43.74
47.27
51.08
55.20
59.66
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
15.95
17.47
19.12
20.91
22.85
24.95
27.22
29.68
32.34
35.21
38.32
41.68
45.31
49.24
53.50
58.10
63.07
68.46
74.30
80.62
87.48
20
20.98
23.15
25.52
28.10
30.90
33.95
37.27
40.88
44.80
49.05
53.67
58.68
64.13
70.03
76.45
83.40
90.96
99.16
108.08
117.76
128.28
40
24.64
27.30
30.21
33.40
36.87
40.66
44.79
49.30
54.20
59.54
65.36
71.69
78.57
86.05
94.20
103.05
112.68
123.16
134.56
146.97
160.48
60
27.61
30.69
34.06
37.75
41.79
46.21
51.03
56.30
62.05
68.33
75.17
82.62
90.75
99.60
109.24
119.74
131.18
143.64
157.21
172.00
188.12
80
30.16
33.60
37.37
41.51
46.05
51.02
56.46
62.41
68.92
76.02
83.78
92.24
101.48
111.56
122.54
134.52
147.59
161.83
177.36
194.31
212.79
100
39.70
44.53
49.88
55.77
62.27
69.43
77.31
85.96
95.46
105.90
117.33
129.87
143.61
158.65
175.11
193.13
212.84
234.40
257.99
283.80
312.03
200
46.61
52.51
59.05
66.29
74.30
83.14
92.90
103.66
115.51
128.55
142.89
158.65
175.95
194.94
215.78
238.62
263.67
291.13
321.22
354.20
390.35
300
52.24
59.02
66.56
74.93
84.21
94.48
105.84
118.39
132.24
147.51
164.33
182.85
203.23
225.62
250.23
277.26
306.94
339.52
375.28
414.51
457.57
400
57.06
64.62
73.04
82.40
92.80
104.33
117.11
131.24
146.87
164.12
183.16
204.14
227.26
252.71
280.71
311.50
345.34
382.53
423.39
468.28
517.58
500
From “Design of Pile Foundations,” by A. S. Vesic. SYNTHESIS OF HIGHWAY PRACTICE by AMERICAN ASSOCIATION OF STATE HIGHWAY AND
TRANSPORT. Copyright 1969 by TRANSPORTATION RESEARCH BOARD. Reproduced with permission of TRANSPORTATION RESEARCH BOARD in
the format Textbook via Copyright Clearance Center.
10
f9
I rr
Table 11.7 Bearing Capacity Factors N s* Based on the Theory of Expansion of Cavities
11.9 Coyle and Castello’s Method for Estimating Qp in Sand
563
Table 11.8 Variation of N c* with Irr for
f 5 0 Condition based on Vesic’s Theory
Irr
Nc*
10
20
40
60
80
100
200
300
400
500
6.97
7.90
8.82
9.36
9.75
10.04
10.97
11.51
11.89
12.19
For f 5 0,
Ir 5
Es
3cu
(11.34)
O’ Neill and Reese (1999) suggested the following approximate relationships for Ir
and the undrained cohesion, cu.
cu
pa
Ir
0.24
0.48
ⱖ0.96
50
150
250⫺300
Note: pa ⫽ atmospheric pressure
< 100 kN>m2.
The preceding values can be approximated as
Ir 5 347a
11.9
cu
b 2 33 # 300
pa
(11.35)
Coyle and Castello’s Method for Estimating Qp in Sand
Coyle and Castello (1981) analyzed 24 large-scale field load tests of driven piles in sand.
On the basis of the test results, they suggested that, in sand,
Qp 5 qrN q*A p
where
qr 5 effective vertical stress at the pile tip
N *q 5 bearing capacity factor
Figure 11.15 shows the variation of N *q with L>D and the soil friction angle fr.
(11.36)
564 Chapter 11: Pile Foundations
10
Bearing capacity factor, N*q
20
40 60 80 100
200
0
10
Embedment ratio, L/D
20
30
40
50
60
32° 36°
40°
30°
34° 38°
Figure 11.15 Variation of N q* with L>D
(Redrawn after Coyle and Costello, 1981)
70
Example 11.1
Consider a 15-m long concrete pile with a cross section of 0.45 m 3 0.45 m fully embedded in sand. For the sand, given: unit weight, g 5 17kN>m3; and soil friction angle,
f r 5 35°. Estimate the ultimate point Qp with each of the following:
a.
b.
c.
d.
Meyerhof’s method
Vesic’s method
The method of Coyle and Castello
based on the results of parts a, b, and c, adopt a value for Qp
Solution
Part a
From Eqs. (11.16) and (11.17),
Qp 5 A pq rN q* # A p (0.5paN q* tan fr )
For fr 5 35°, the value of N q* < 143 (Table 11.5). Also, q r 5 gL 5 (17) (15)
5 255kN>m2.
Thus,
A pq rN *q 5 (0.45 3 0.45) (255) (143) < 7384kN
Again,
A p (0.5paN *q tan fr ) 5 (0.45 3 0.45) 3(0.5) (100) (143) ( tan 35)4 < 1014 kN
Hence, Qp 5 1014 kN.
Part b
From Eq. (11.19),
11.9 Coyle and Castello’s Method for Estimating Qp in Sand
565
Qp 5 A psor N s*
sor 5 c
1 1 2(1 2 sin fr )
1 1 2(1 2 sin 35)
d qr 5 a
b (17 3 15)
3
3
5 139.96kN>m2
From Eq. (11.26),
Es
5m
pa
Assume m < 250 (medium sand). So,
Es 5 (250) (100) 5 25,000 kN>m2
From Eq. (11.27),
ms 5 0.1 1 0.3a
fr 2 25
35 2 25
b 5 0.1 1 0.3a
b 5 0.25
20
20
From Eq. (11.28),
D 5 0.005a1 2
fr 2 25
qr
35 2 25
17 3 15
ba
b 5 0.005a1 2
ba
b 5 0.0064
pa
20
20
100
From Eq. (11.25),
Ir 5
Es
25,000
5
5 56
2(1 1 ms )q r tan fr
(2) (1 1 0.25) (17 3 15) ( tan 35)
From Eq. (11.24),
Irr 5
Ir
56
5
5 41.2
1 1 Ir D
1 1 (56) (0.0064)
From Table 11.7, for fr 5 35° and Irr 5 41.2, the value of N *s < 55. Hence,
Qp 5 A psor N s* 5 (0.45 3 0.45) (139.96) (55) < 1559 kN
Part c
From Eq. (11.36),
Qp 5 q rN *q A p
L
15
5
5 33.3
D
0.45
For fr 5 35° and L>D 5 33.3, the value of N *q is about 48 (Figure 11.15). Thus,
Qp 5 q rN *q A p 5 (15 3 17) (48) (0.45 3 0.45) < 2479 kN
Part d
It appears that Qp obtained from the method of Coyle and Castello is too large. Thus,
the average of the results from parts a and b is
566 Chapter 11: Pile Foundations
1014 1 1559
5 1286.5 kN
2
Use Qp 5 1250 kN.
■
Example 11.2
Consider a pipe pile (flat driving point—see Figure 11.2d) having an outside diameter
of 406 mm. The embedded length of the pile in layered saturated clay is 30 m.
The following are the details of the subsoil:
Depth from
ground surface
(m)
Saturated unit
weight,
g(kN>m3 )
cu (kN>m2 )
0–5
5–10
10–30
18
18
19.6
30
30
100
The groundwater table is located at a depth of 5 m from the ground surface. Estimate
Qp by using
a. Meyerhof’s method
b. Vesic’s method
Solution
Part a
From Eq. (11.18),
Qp 5 9cuA p
The tip of the pile is resting on a clay with cu 5 100 kN>m2. So,
p
406 2
Qp 5 (9) (100) c a b a
b d 5 116.5 kN
4
1000
Part b
From Eq. (11.31),
Qp 5 A pcuN *c
From Eq. (11.35),
Ir 5 Irr 5 347a
cu
100
b 2 33 5 347a
b 2 33 5 314
pa
100
So we use Irr 5 300.
From Table 11.8 for Irr 5 300, the value of N c* 5 11.51. Thus,
p
406 2
Qp 5 A pcuN *c 5 c a b a
b d (100) (11.51) 5 149.0 kN
4 1000
11.10 Correlations for Calculating Qp with SPT and CPT Results
567
Note: The average value of Qp is
116.5 1 149.0
< 133 kN
2
11.10
■
Correlations for Calculating Qp with SPT
and CPT Results
On the basis of field observations, Meyerhof (1976) also suggested that the ultimate point
resistance qp in a homogeneous granular soil (L 5 Lb ) may be obtained from standard
penetration numbers as
L
qp 5 0.4paN60
# 4paN60
(11.37)
D
where
N60 5 the average value of the standard penetration number near the pile point (about
10D above and 4D below the pile point)
pa 5 atmospheric pressure ( < 100 kN>m2 or 2000 lb>ft 2 )
Briaud et al. (1985) suggested the following correlation for qp in granular soil with
the standard penetration resistance N60.
qp 5 19.7pa (N60 ) 0.36
Meyerhof (1956) also suggested that
qp < qc (in granular soil)
where qc 5 cone penetration resistance.
(11.38)
(11.39)
Example 11.3
Consider a concrete pile that is 0.305 m 3 0.305 m in cross section in sand. The pile is
15.2 m long. The following are the variations of N60 with depth.
Depth below ground surface (m)
1.5
3.0
4.5
6.0
7.5
9.0
10.5
12.0
13.5
15.0
16.5
18.0
19.5
21.0
N 60
8
10
9
12
14
18
11
17
20
28
29
32
30
27
568 Chapter 11: Pile Foundations
a. Estimate Qp using Eq. (11.37).
b. Estimate Qp using Eq. (11.38).
Solution
Part a
The tip of the pile is 15.2 m below the ground surface. For the pile, D 5 0.305 m. The
average of N60 10D above and about 5D below the pile tip is
N60 5
17 1 20 1 28 1 29
5 23.5 < 24
4
From Eq. (11.37)
Qp 5 A p (qp ) 5 A p c0.4paN60 a
A p c0.4paN60 a
L
b d # A p (4paN60 )
D
L
15.2
b d 5 (0.305 3 0.305) c (0.4) (100) (24) a
b d 5 4450.6 kN
D
0.305
A p (4paN60 ) 5 (0.305 3 0.305) 3(4) (100) (24)4 5 893 kN
Thus, Qp 5 893 kN
Part b
From Eq. (11.38),
Qp 5 A pqp 5 A p 319.7pa (N60 ) 0.364 5 (0.305 3 0.305) 3(19.7) (100) (24) 0.364
5 575.4 kN
11.11
■
Frictional Resistance (Qs) in Sand
According to Eq. (11.14), the frictional resistance
Qs 5 Sp DLf
The unit frictional resistance, f, is hard to estimate. In making an estimation of f, several
important factors must be kept in mind:
1. The nature of the pile installation. For driven piles in sand, the vibration caused
during pile driving helps densify the soil around the pile. The zone of sand densification may be as much as 2.5 times the pile diameter, in the sand surrounding
the pile.
2. It has been observed that the nature of variation of f in the field is approximately
as shown in Figure 11.16. The unit skin friction increases with depth more or
11.11 Frictional Resistance (Qs) in Sand
569
Unit
frictional
resistance, f
D
L
z
f
Ko
L
L
Depth
(a)
(b)
Figure 11.16 Unit frictional resistance for piles in sand
less linearly to a depth of Lr and remains constant thereafter. The magnitude of
the critical depth Lr may be 15 to 20 pile diameters. A conservative estimate
would be
Lr < 15D
(11.40)
3. At similar depths, the unit skin friction in loose sand is higher for a highdisplacement pile, compared with a low-displacement pile.
4. At similar depths, bored, or jetted, piles will have a lower unit skin friction compared with driven piles.
Taking into account the preceding factors, we can give the following approximate
relationship for f (see Figure 11.16):
For z 5 0 to Lr,
f 5 Ksor tan d r
(11.41)
f 5 fz5Lr
(11.42)
and for z 5 Lr to L,
In these equations,
K 5 effective earth pressure coefficient
sor 5 effective vertical stress at the depth under consideration
dr 5 soil-pile friction angle
In reality, the magnitude of K varies with depth; it is approximately equal to the
Rankine passive earth pressure coefficient, Kp , at the top of the pile and may be less than
570 Chapter 11: Pile Foundations
the at-rest pressure coefficient, Ko , at a greater depth. Based on presently available
results, the following average values of K are recommended for use in Eq. (11.41):
K
Pile type
<Ko 5 1 2 sin fr
<Ko 5 1 2 sin fr to 1.4Ko 5 1.4(1 2 sin fr)
<Ko 5 1 2 sin fr to 1.8Ko 5 1.8(1 2 sin fr)
Bored or jetted
Low-displacement driven
High-displacement driven
The values of dr from various investigations appear to be in the range from 0.5fr
to 0.8fr.
Based on load test results in the field, Mansur and Hunter (1970) reported the following average values of K.
H-pilesccK 5 1.65
Steel pipe pilesccK 5 1.26
Precast concrete pilesccK 5 1.5
Coyle and Castello (1981), in conjunction with the material presented in Section
11.9, proposed that
Qs 5 favpL 5 (Ksor tan dr )pL
(11.43)
where
sor 5 average effective overburden pressure
dr 5 soil–pile friction angle 5 0.8fr
The lateral earth pressure coefficient K, which was determined from field observations, is
shown in Figure 11.17. Thus, if that figure is used,
Qs 5 Ksor tan(0.8fr)pL
(11.44)
Correlation with Standard Penetration Test Results
Meyerhof (1976) indicated that the average unit frictional resistance, fav , for highdisplacement driven piles may be obtained from average standard penetration resistance values as
fav 5 0.02pa (N60 )
(11.45)
where
(N60 ) 5 average value of standard penetration resistance
pa 5 atmospheric pressure (<100 kN>m2 or 2000 lb>ft 2 )
For low-displacement driven piles
fav 5 0.01pa (N60 )
(11.46)
11.11 Frictional Resistance (Qs) in Sand
0.15 0.2
0
Earth pressure coefficient, K
1.0
2
571
5
30°
31°
5
32°
33°
10
35°
Embedment ratio, L /D
34°
36°
15
20
25
30
Figure 11.17 Variation of K
with L>D (Redrawn after Coyle
and Castello, 1981)
35
36
Briaud et al. (1985) suggested that
fav < 0.224pa (N60 ) 0.29
(11.47)
Qs 5 pLfav
(11.48)
Thus,
Correlation with Cone Penetration Test Results
Nottingham and Schmertmann (1975) and Schmertmann (1978) provided correlations for
estimating Qs using the frictional resistance (fc ) obtained during cone penetration tests.
According to this method
f 5 arfc
(11.49)
The variations of ar with z>D for electric cone and mechanical cone penetrometers are
shown in Figures 11.18 and 11.19, respectively. We have
Qs 5 Sp(DL)f 5 Sp(DL)arfc
(11.50)
572 Chapter 11: Pile Foundations
3.0
Schmertmann (1978);
Nottingham and Schmertmann (1975)
Steel
pile
2.0
Timber pile
Concrete pile
1.0
0
0
10
20
L /D
30
40
Figure 11.18 Variation of ar with embedment ratio for pile in sand: electric cone penetrometer
2.0
Schmertmann (1978);
Nottingham and Schmertmann (1975)
1.5
Steel
pile
Timber
pile
1.0
Concrete
pile
0.5
0
0
10
20
L /D
30
40
Figure 11.19 Variation of ar with embedment ratio for piles in sand: mechanical cone
penetrometer
Example 11.4
Refer to the pile described in Example 11.3. Estimate the magnitude of Qs for the pile.
a. Use Eq. (11.45).
b. Use Eq. (11.47).
11.11 Frictional Resistance (Qs) in Sand
573
c. Considering the results in Example 11.3, determine the allowable load-carrying
capacity of the pile based on Meyerhof’s method and Briaud’s method. Use a
factor of safety, FS 5 3.
Solution
The average N60 value for the sand for the top 15.2 m is
N60 5
8 1 10 1 9 1 12 1 14 1 18 1 11 1 17 1 20 1 28
5 14.7 < 15
10
Part a
From Eq. (11.45),
fav 5 0.02pa (N60 ) 5 (0.02) (100) (15) 5 30 kN>m2
Qs 5 pLfav 5 (4 3 0.305) (15.2) (30) 5 556.2 kN
Part b
From Eq. (11.47),
fav 5 0.224pa (N60 ) 0.29 5 (0.224) (100) (15) 0.29 5 49.13 kN>m2
Qs 5 pLfav 5 (4 3 0.305) (15.2) (49.13) 5 911.1 kN
Part c
Qp 1 Qs
893 1 556.2
5 483 kN
3
575.4 1 911.1
Briaud’s method: Qall 5
5
5 495.5 kN
FS
3
So the allowable pile capacity may be taken to be about 490 kN.
Meyerhof’s method: Qall 5
FS
Qp 1 Qs
5
■
Example 11.5
Refer to Example 11.1. For the pile, estimate the frictional resistance Qs
a. Based on Eqs. (11.41) and (11.42). Use K 5 1.3 and dr 5 0.8fr.
b. Based on Eq. (11.44).
c. Using the results of Part d of Example 11.1, estimate the allowable bearing
capacity of the pile. Use FS 5 3.
Solution
Part a
From Eq. (11.40), Lr 5 15D 5 (15) (0.45) 5 6.75m. Refer to Eq. (11.41):
At z 5 0:
sor 5 0
f50
At z 5 6.75 m:
sor 5 (6.75) (17) 5 114.75 kN>m2
574 Chapter 11: Pile Foundations
So
f 5 Ksor tan d 5 (1.3) (114.75) 3tan (0.8 3 35)4 5 79.3 kN>m3
Thus,
Qs 5
(fz50 1 fz56.75m )
pLr 1 fz56.75m p(L 2 Lr )
2
0 1 79.3
5a
b (4 3 0.45) (6.75) 1 (79.3) (4 3 0.45) (15 2 6.75)
2
5 481.75 1 1177.61 5 1659.36 kN < 1659 kN
Part b
From Eq. (11.44),
Qs 5 Ksor tan (0.8fr )pL
(15) (17)
sor 5
5 127.5 kN>m2
2
L
15
5
5 33.3; fr 5 35°
D
0.45
From Figure 11.17, K 5 0.93
Qs 5 (1.3) (127.5) tan3 (0.8 3 35)4 (4 3 0.45) (15) 5 2380 kN
Part c
The average value of Qs from parts a and b is
Qs(average) 5
1659 1 2380
5 2019.5 < 2020 kN 2 USE
2
From part d of Example 11.1, Qp 5 1250 kN. Thus,
Qall 5
Qp 1 Qs
FS
5
1250 1 2020
5 1090 kN
3
■
Example 11.6
Consider an 18-m long concrete pile (cross section: 0.305 m 3 0.305 m) fully embedded in a sand layer. For the sand layer, the following is an approximation of the cone
penetration resistance qc (mechanical cone) and the frictional resistance fc with depth.
Estimate the allowable load that the pile can carry. Use FS 5 3.
Depth from
ground surface (m)
qc (kN>m2 )
fc (kN>m2 )
0–5
5–15
15–25
3040
4560
9500
73
102
226
11.12 Frictional (Skin) Resistance in Clay
575
Solution
Qu 5 Qp 1 Qs
From Eq. (11.39),
qp < qc
At the pile tip (i.e., at a depth of 18 m), qc < 9500 kN>m2. Thus,
Qp 5 A pqc 5 (0.305 3 0.305) (9500) 5 883.7 kN
To determine Qs, the following table can be prepared. (Note: L>D 5 18>0.305 5 59.)
Depth from
ground surface (m)
0–5
5–15
15–18
⌬L (m)
fc (kN/m2)
a9
(Figure 11.19)
p⌬L␣9fc (kN)
5
10
3
73
102
226
0.44
0.44
0.44
195.9
547.5
363.95
Qs 5 1107.35 kN
Hence,
Qu 5 Qp 1 Qs 5 883.7 1 1107.35 5 1991.05 kN
Qu
1991.05
Qall 5
5
5 663.68 < 664 kN
FS
3
11.12
■
Frictional (Skin) Resistance in Clay
Estimating the frictional (or skin) resistance of piles in clay is almost as difficult a task as
estimating that in sand (see Section 11.11), due to the presence of several variables that cannot easily be quantified. Several methods for obtaining the unit frictional resistance of piles
are described in the literature. We examine some of them next.
l Method
This method, proposed by Vijayvergiya and Focht (1972), is based on the assumption that
the displacement of soil caused by pile driving results in a passive lateral pressure at any
depth and that the average unit skin resistance is
fav 5 l(sor 1 2cu )
where
sor 5 mean effective vertical stress for the entire embedment length
cu 5 mean undrained shear strength (f 5 0)
(11.51)
576 Chapter 11: Pile Foundations
Table 11.9 Variation of l with pile embedment
length, L
Embedment
length, L (m)
l
0
5
10
15
20
25
30
35
40
50
60
70
80
90
0.5
0.336
0.245
0.200
0.173
0.150
0.136
0.132
0.127
0.118
0.113
0.110
0.110
0.110
The value of l changes with the depth of penetration of the pile. (See Table 11.9.) Thus, the total frictional resistance may be calculated as
Qs 5 pLfav
Care should be taken in obtaining the values of sor and cu in layered soil. Figure 11.20 helps
explain the reason. Figure 11.20a shows a pile penetrating three layers of clay. According to
Figure 11.20b, the mean value of cu is (cu(1)L1 1 cu(2)L2 1 c )>L. Similarly, Figure 11.20c
shows the plot of the variation of effective stress with depth. The mean effective stress is
A1 1 A2 1 A3 1 c
(11.52)
sor 5
L
where A 1 , A 2 , A 3 , c 5 areas of the vertical effective stress diagrams.
Undrained
cohesion, cu
L1
L3
Area A2
cu(2)
L2
L
Area A1
cu(1)
Area A3
cu(3)
(a)
Depth
(b)
Vertical
effective
stress, o
Depth
(c)
Figure 11.20 Application of l method in layered soil
11.12 Frictional (Skin) Resistance in Clay
577
a Method
According to the a method, the unit skin resistance in clayey soils can be represented by
the equation
f 5 acu
(11.53)
where a 5 empirical adhesion factor. The approximate variation of the value of a is
shown in Table 11.10. It is important to realize that the values of a given in Table 11.10
may vary somewhat, since a is actually a function of vertical effective stress and the
undrained cohesion. Sladen (1992) has shown that
a 5 Ca
sor 0.45
b
cu
(11.54)
where
sor 5 average vertical effective stress
C < 0.4 to 0.5 for bored piles and $ 0.5 for driven piles
The ultimate side resistance can thus be given as
Qs 5 Sfp DL 5 S acup DL
Table 11.10 Variation of a (interpolated values based on Terzaghi, Peck
and Mesri, 1996)
cu
pa
# 0.1
0.2
0.3
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.4
2.8
a
1.00
0.92
0.82
0.74
0.62
0.54
0.48
0.42
0.40
0.38
0.36
0.35
0.34
0.34
Note: pa 5 atmospheric pressure
< 100 kN>m2
(11.55)
578 Chapter 11: Pile Foundations
b Method
When piles are driven into saturated clays, the pore water pressure in the soil around the piles
increases. The excess pore water pressure in normally consolidated clays may be four to six
times cu . However, within a month or so, this pressure gradually dissipates. Hence, the unit
frictional resistance for the pile can be determined on the basis of the effective stress parameters of the clay in a remolded state (cr 5 0). Thus, at any depth,
f 5 bsor
(11.56)
where
sor 5 vertical effective stress
b 5 K tan fRr
fRr 5 drained friction angle of remolded clay
K 5 earth pressure coefficient
(11.57)
Conservatively, the magnitude of K is the earth pressure coefficient at rest, or
K 5 1 2 sin fRr
(for normally consolidated clays)
(11.58)
and
K 5 (1 2 sin fRr )"OCR
(for overconsolidated clays)
(11.59)
where OCR 5 overconsolidation ratio.
Combining Eqs. (11.56), (11.57), (11.58), and (11.59), for normally consolidated
clays yields
f 5 (1 2 sin fRr ) tan fRr sor
(11.60)
and for overconsolidated clays,
f 5 (1 2 sin fRr )tan fRr "OCR sor
(11.61)
With the value of f determined, the total frictional resistance may be evaluated as
Qs 5 Sfp DL
Correlation with Cone Penetration Test Results
Nottingham and Schmertmann (1975) and Schmertmann (1978) found the correlation for
unit skin friction in clay (with f 5 0) to be
f 5 arfc
(11.62)
The variation of ar with the frictional resistance fc is shown in Figure 11.21. Thus,
Qs 5 Sfp(DL) 5 Sarfcp(DL)
(11.63)
11.13 Point Bearing Capacity of Piles Resting on Rock
579
1.5
Nottingham and Schmertmann (1975);
Schmertmann (1978)
1.25
1.0
0.75
Concrete and
timber piles
0.5
0.25
Steel piles
0
0
0.5
1.0
fc
pa
1.5
2.0
Figure 11.21 Variation of ar with fc>pa for piles in clay (pa 5 atmospheric pressure
<100 kN>m2)
11.13
Point Bearing Capacity of Piles Resting on Rock
Sometimes piles are driven to an underlying layer of rock. In such cases, the engineer must
evaluate the bearing capacity of the rock. The ultimate unit point resistance in rock
(Goodman, 1980) is approximately
qp 5 qu (Nf 1 1)
(11.64)
where
Nf 5 tan2 (45 1 fr>2)
qu 5 unconfined compression strength of rock
fr 5 drained angle of friction
The unconfined compression strength of rock can be determined by laboratory tests on
rock specimens collected during field investigation. However, extreme caution should
be used in obtaining the proper value of qu , because laboratory specimens usually are
small in diameter. As the diameter of the specimen increases, the unconfined compression strength decreases—a phenomenon referred to as the scale effect. For specimens
larger than about 1 m in diameter, the value of qu remains approximately constant.
There appears to be a fourfold to fivefold reduction of the magnitude of qu in this
process. The scale effect in rock is caused primarily by randomly distributed large and
580 Chapter 11: Pile Foundations
small fractures and also by progressive ruptures along the slip lines. Hence, we always
recommend that
qu(design) 5
qu(lab)
5
(11.65)
Table 11.11 lists some representative values of (laboratory) unconfined compression
strengths of rock. Representative values of the rock friction angle fr are given in Table 11.12.
A factor of safety of at least 3 should be used to determine the allowable point
bearing capacity of piles. Thus,
Qp(all) 5
3qu(design) (Nf 1 1)4A p
FS
(11.66)
Table 11.11 Typical Unconfined Compressive
Strength of Rocks
Type of rock
qu
MN , m2
Sandstone
Limestone
Shale
Granite
Marble
70–140
105–210
35–70
140–210
60–70
Table 11.12 Typical Values of Angle of Friction fr
of Rocks
Type of rock
Sandstone
Limestone
Shale
Granite
Marble
Angle of friction, f9 (deg)
27–45
30–40
10–20
40–50
25–30
Example 11.7
Refer to the pile in saturated clay shown in Figure 11.22. For the pile,
a. Calculate the skin resistance (Qs ) by (1) the a method, (2) the l method, and
(3) the b method. For the b method, use fRr 5 30° for all clay layers. The top
10 m of clay is normally consolidated. The bottom clay layer has an
OCR 5 2. (Note: diameter of pile ⫽ 406 mm)
b. Using the results of Example 11.2, estimate the allowable pile capacity (Qall ) .
Use FS 5 4.
11.13 Point Bearing Capacity of Piles Resting on Rock
581
o (kN/m2)
Saturated clay
cu(1) 30 kN/m2
18 kN/m3
5m
A1 225
Groundwater
table
cu(1) 30 kN/m2
Clay
18 kN/m3
5m
5m
90
A2 552.38
130.95
10 m
z
Clay
cu(2) 100 kN/m2
sat 19.6 kN/m3
20 m
A3 4577
30 m
(a)
326.75
(b)
Figure 11.22 Estimation of the load bearing capacity of a driven-pipe pile
Solution
Part a
(1) From Eq. (11.55),
Qs 5 SacupDL
[Note: p 5 p(0.406) 5 1.275m] Now the following table can be prepared.
Depth
(m)
DL
(m)
cu
(kN/m2)
a
(Table 11.10)
␣cup DL
(kN)
0–5
5–10
10–30
5
5
20
30
30
100
0.82
0.82
0.48
156.83
156.83
1224.0
Qs < 1538 kN
(2) From Eq. 11.51, fav 5 l(sor 1 2cu ). Now, the average value of cu is
cu(1) (10) 1 cu(2) (20)
30
5
(30) (10) 1 (100) (20)
5 76.7 kN>m2
30
To obtain the average value of sor , the diagram for vertical effective stress variation with
depth is plotted in Figure 11.22b. From Eq. (11.52),
sor 5
A1 1 A2 1 A3
225 1 552.38 1 4577
5
5 178.48 kN>m2
L
30
582 Chapter 11: Pile Foundations
From Table 11.9, the magnitude of l is 0.136. So
fav 5 0.1363178.48 1 (2) (76.7) 4 5 45.14 kN>m2
Hence,
Qs 5 pLfav 5 p(0.406) (30) (45.14) 5 1727 kN
(3) The top layer of clay (10 m) is normally consolidated, and fRr 5 30°. For
z 5 0–5 m, from Eq. (11.60), we have
fav(1) 5 (1 2 sin fRr ) tan fRr sor
5 (1 2 sin 30°) (tan 30°) ¢
0 1 90
≤ 5 13.0 kN>m2
2
Similarly, for z 5 5–10 m.
fav(2) 5 (1 2 sin 30°) (tan 30°) ¢
90 1 130.95
≤ 5 31.9 kN>m2
2
For z 5 10–30 m from Eq. (11.61),
fav 5 (1 2 sin fRr )tan fRr "OCR sor
For OCR 5 2,
fav(3) 5 (1 2 sin 30°) (tan 30°)"2¢
130.95 1 326.75
≤ 5 93.43 kN>m2
2
So,
Qs 5 p3fav(1) (5) 1 fav(2) (5) 1 fav(3) (20)4
5 (p) (0.406) 3(13) (5) 1 (31.9) (5) 1 (93.43) (20)4 5 2670 kN
Part b
Qu 5 Qp 1 Qs
From Example 11.2,
Qp <
116.5 1 149
< 133 kN
2
Again, the values of Qs from the a method and l method are close. So,
Qs <
Qall
1538 1 1727
5 1632.5 kN
2
Qu
133 1 1632.5
5
5
5 441.4 kN < 441 kN
FS
4
■
11.14 Pile Load Tests
583
Example 11.8
A concrete pile 305 mm 3 305 mm in cross section is driven to a depth of 20 m below
the ground surface in a saturated clay soil. A summary of the variation of frictional resistance fc obtained from a cone penetration test is as follows:
Depth
(m)
Friction resistance,
fc (kg/cm2)
0–6
6–12
12–20
0.35
0.56
0.72
Estimate the frictional resistance Qs for the pile.
Solution
We can prepare the following table:
Depth
(m)
fc
(kN/m2)
a9
(Figure 11.21)
DL
(m)
a9fc p(DL)
[Eq. (11.63)] (kN)
0 –6
6 –12
12–20
34.34
54.94
70.63
0.84
0.71
0.63
6
6
8
211.5
285.5
434.2
(Note: p 5 (4) (0.305) 5 1.22 m)
Thus,
Qs 5 g arfc p(DL) 5 931 kN
11.14
■
Pile Load Tests
In most large projects, a specific number of load tests must be conducted on piles. The
primary reason is the unreliability of prediction methods. The vertical and lateral loadbearing capacity of a pile can be tested in the field. Figure 11.23a shows a schematic
diagram of the pile load arrangement for testing axial compression in the field. The load
is applied to the pile by a hydraulic jack. Step loads are applied to the pile, and sufficient
time is allowed to elapse after each load so that a small amount of settlement occurs. The
settlement of the pile is measured by dial gauges. The amount of load to be applied for
each step will vary, depending on local building codes. Most building codes require that
each step load be about one-fourth of the proposed working load. The load test should be
carried out to at least a total load of two times the proposed working load. After the desired
pile load is reached, the pile is gradually unloaded.
584 Chapter 11: Pile Foundations
Beam
Hydraulic
jack
Dial
gauge
Reference
beam
Anchor
pile
Test
pile
(a)
Q1
Q2
st(1)
Load, Q
Load, Q
st(2)
Qu
Loading
se(1)
Qu
se(2)
Unloading
1
Settlement
Net settlement, snet
(b)
2
(c)
Figure 11.23 (a) Schematic diagram of pile load test arrangement (b) plot
of load against total settlement; (c) plot of load against net settlement
Figure 11.23b shows a load–settlement diagram obtained from field loading and
unloading. For any load Q, the net pile settlement can be calculated as follows:
When Q 5 Q1 ,
Net settlement, snet(1) 5 st(1) 2 se(1)
When Q 5 Q2 ,
Net settlement, snet(2) 5 st(2) 2 se(2)
(
where
snet 5 net settlement
se 5 elastic settlement of the pile itself
st 5 total settlement
These values of Q can be plotted in a graph against the corresponding net settlement, snet , as
shown in Figure 11.23c. The ultimate load of the pile can then be determined from the graph.
11.14 Pile Load Tests
585
Pile settlement may increase with load to a certain point, beyond which the load–settlement
curve becomes vertical. The load corresponding to the point where the curve of Q versus snet
becomes vertical is the ultimate load, Qu , for the pile; it is shown by curve 1 in Figure 11.23c.
In many cases, the latter stage of the load–settlement curve is almost linear, showing a large
degree of settlement for a small increment of load; this is shown by curve 2 in the figure. The
ultimate load, Qu , for such a case is determined from the point of the curve of Q versus snet
where this steep linear portion starts.
One of the methods to obtain the ultimate load Qu from the load-settlement plot is
that proposed by Davisson (1973). Davisson’s method is used more often in the field and
is described here. Referring to Figure 11.24, the ultimate load occurs at a settlement
level (su ) of
su (mm) 5 0.012Dr 1 0.1a
QuL
D
b1
Dr
ApEp
where
Qu is in kN
D is in mm
Dr 5 reference pile diameter or width (5 300mm)
L 5 pile length (mm)
A p 5 area of pile cross section (mm2 )
Ep 5 Young’s modulus of pile material (kN>mm2 )
Qu
Load, Q (kN)
0.12 Dr+
0.1 D/Dr
QuL
AE
Eq. (11.67)
Settlement, s (mm)
Figure 11.24 Davisson’s method for determination of Qu
(11.67)
586 Chapter 11: Pile Foundations
The application of this procedure is shown in Example 11.9.
The load test procedure just described requires the application of step loads on
the piles and the measurement of settlement and is called a load-controlled test.
Another technique used for a pile load test is the constant-rate-of-penetration test,
wherein the load on the pile is continuously increased to maintain a constant rate of
penetration, which can vary from 0.25 to 2.5 mm>min (0.01 to 0.1 in.>min). This test
gives a load–settlement plot similar to that obtained from the load-controlled test.
Another type of pile load test is cyclic loading, in which an incremental load is repeatedly applied and removed.
In order to conduct a load test on piles, it is important to take into account the time
lapse after the end of driving (EOD). When piles are driven into soft clay, a certain zone
surrounding the clay becomes remolded or compressed, as shown in Figure 11.25a. This
results in a reduction of undrained shear strength, cu (Figure 11.25b). With time, the loss
of undrained shear strength is partially or fully regained. The time lapse may range from
30 to 60 days.
For piles driven in dilative (dense to very dense) saturated fine sands, relaxation is
possible. Negative pore water pressure, if developed during pile driving, will dissipate over
time, resulting in a reduction in pile capacity with time after the driving operation is completed. At the same time, excess pore water pressure may be generated in contractive fine
0.5 D
1.5 D
Pile
Remolded
zone
Compressed
zone
D diameter
0.5 D
1.5 D
(a)
cu
Remolded
zone
Sometime
after
driving
Compressed
zone
Intact
zone
Immediately
after
driving
Distance from pile
(b)
Figure 11.25 (a) Remolded or compacted zone around a pile driven into soft clay;
(b) Nature of variation of undrained shear strength (cu ) with time around a pile driven into
soft clay
Intact
zone
11.14 Pile Load Tests
587
sands during pile driving. The excess pore water pressure will dissipate over time, which
will result in greater pile capacity.
Several empirical relationships have been developed to predict changes in pile
capacity with time.
Skov and Denver (1988)
Skov and Denver proposed the equation
t
Qt 5 QOED cA log a b 1 1d
to
where
(11.68)
Qt 5 pile capacity t days after the end of driving
QOED 5 pile capacity at the end of driving
t 5 time, in days
For sand, A 5 0.2 and to 5 0.5 days; for clay, A 5 0.6 and to 5 1.0 days.
Guang-Yu (1988)
According to Guang-Yu,
Q14 5 (0.375St 1 1)QOED
where
(applicable to clay soil)
(11.69)
Q14 5 pile capacity 14 days after pile driving
St 5 sensitivity of clay
Svinkin (1996)
Svinkin suggests the relationship
Qt 5 1.4QOEDt0.1 (upper limit for sand)
Qt 5 1.025QOEDt0.1 (lower limit for sand)
(11.70)
(11.71)
where t 5 time after driving, in days
Example 11.9
Figure 11.26 shows the load test results of a 20-m long concrete pile (406 mm 3 406 mm)
embedded in sand. Using Davisson’s method, determine the ultimate load Qu. Given:
Ep 5 30 3 106 kN>m2.
Solution
From Eq. (11.67),
su 5 0.012Dr 1 0.1a
QuL
D
b1
Dr
A pEp
588 Chapter 11: Pile Foundations
Dr 5 300 mm, D 5 406 mm, L 5 20 m 5 20,000 mm, A p 5 406 mm 3 406 mm 5
164,836 mm2, and Ep 5 30 3 106 kN>m2. Hence,
su 5 (0.012) (300) 1 (0.1) a
(Qu ) (20,000)
406
b1
300
(30) (164,836)
5 3.6 1 0.135 1 0.004Qu 5 3.735 1 0.004Qu
The line su (mm) 5 3.735 1 0.004Qu is drawn in Figure 11.26. The intersection of this
line with the load-settlement curve gives the failure load Qu 5 1640 kN.
800
Qu = 1460 kN
1600
2400
Q (kN)
3.735 mm
5
10
15
20
Settlement, s (mm)
Figure 11.26
11.15
■
Elastic Settlement of Piles
The total settlement of a pile under a vertical working load Qw is given by
se 5 se(1) 1 se(2) 1 se(3)
(11.72)
11.15 Elastic Settlement of Piles
589
where
se(1) 5 elastic settlement of pile
se(2) 5 settlement of pile caused by the load at the pile tip
se(3) 5 settlement of pile caused by the load transmitted along the pile shaft
If the pile material is assumed to be elastic, the deformation of the pile shaft can be evaluated, in accordance with the fundamental principles of mechanics of materials, as
se(1) 5
(Qwp 1 jQws )L
A pEp
(11.73)
where
Qwp 5 load carried at the pile point under working load condition
Qws 5 load carried by frictional (skin) resistance under working load condition
A p 5 area of cross section of pile
L 5 length of pile
Ep 5 modulus of elasticity of the pile material
The magnitude of j varies between 0.5 and 0.67 and will depend on the nature of the distribution of the unit friction (skin) resistance f along the pile shaft.
The settlement of a pile caused by the load carried at the pile point may be expressed
in the form:
se(2) 5
qwpD
Es
(1 2 m2s )Iwp
(11.74)
where
D 5 width or diameter of pile
qwp 5 point load per unit area at the pile point 5 Qwp>A p
Es 5 modulus of elasticity of soil at or below the pile point
ms 5 Poisson’s ratio of soil
Iwp 5 influence factor < 0.85
Vesic (1977) also proposed a semi-empirical method for obtaining the magnitude of
the settlement of se(2) . His equation is
se(2) 5
QwpCp
Dqp
where
qp 5 ultimate point resistance of the pile
Cp 5 an empirical coefficient
Representative values of Cp for various soils are given in Table 11.13.
(11.75)
590 Chapter 11: Pile Foundations
Table 11.13 Typical Values of Cp [from Eq. (11.75)]
Type of soil
Driven pile
Bored pile
Sand (dense to loose)
Clay (stiff to soft)
Silt (dense to loose)
0.02–0.04
0.02–0.03
0.03–0.05
0.09–0.18
0.03–0.06
0.09–0.12
From “Design of Pile Foundations,” by A. S. Vesic. SYNTHESIS OF
HIGHWAY PRACTICE by AMERICAN ASSOCIATION OF STATE
HIGHWAY AND TRANSPORT. Copyright 1969 by TRANSPORTATION
RESEARCH BOARD. Reproduced with permission of TRANSPORTATION
RESEARCH BOARD in the format Textbook via Copyright Clearance Center.
The settlement of a pile caused by the load carried by the pile shaft is given by a relation similar to Eq. (11.74), namely,
se(3) 5 ¢
Qws D
≤ (1 2 m2s )Iws
pL Es
(11.76)
where
p 5 perimeter of the pile
L 5 embedded length of pile
Iws 5 influence factor
Note that the term Qws>pL in Eq. (11.76) is the average value of f along the pile shaft. The
influence factor, Iws, has a simple empirical relation (Vesic, 1977):
L
(11.77)
ÅD
Vesic (1977) also proposed a simple empirical relation similar to Eq. (11.75) for
obtaining se(3) :
Iws 5 2 1 0.35
se(3) 5
QwsCs
Lqp
In this equation, Cs 5 an empirical constant 5 (0.93 1 0.16"L>D)Cp
(11.78)
(11.79)
The values of Cp for use in Eq. (11.75) may be estimated from Table 11.13.
Example 11.10
The allowable working load on a prestressed concrete pile 21-m long that has been driven
into sand is 502 kN. The pile is octagonal in shape with D 5 356 mm (see Table 11.3a).
Skin resistance carries 350 kN of the allowable load, and point bearing carries the rest.
Use Ep 5 21 3 106 kN>m2, Es 5 25 3 103 kN>m2, ms 0.35, and j 5 0.62. Determine the
settlement of the pile.
11.16 Laterally Loaded Piles
591
Solution
From Eq. (11.73),
Se(1) 5
(Qwp 1 jQws )L
A pEp
From Table 11.3a for D 5 356 mm, the area of pile cross section. Ap 5 1045 cm2, Also,
perimeter p 5 1.168 m. Given: Qws 5 350 kN, so
Qwp 5 502 2 350 5 152 kN
se(1) 5
3152 1 0.62(350)4 (21)
(0.1045 m2 ) (21 3 106 )
5 0.00353 m 5 3.35 mm
From Eq. (11.74),
qwpD
152
0.356
(1 2 m2s )Iwp 5 ¢
≤¢
≤ (1 2 0.352 ) (0.85)
Es
0.1045 25 3 103
5 0.0155 m 5 15.5 mm
se(2) 5
Again, from Eq. (11.76),
se(3) 5 ¢
Qws
D
≤ ¢ ≤ (1 2 m2s )Iws
pL
Es
L
21
5 2 1 0.35
5 4.69
ÅD
Å 0.356
350
0.356
5B
R¢
≤ (1 2 0.352 ) (4.69)
(1.168) (21)
25 3 103
Iws 5 2 1 0.35
se(3)
5 0.00084 m 5 0.84 mm
Hence, total settlement is
se 5 se(1) 1 se(2) 1 se(3) 5 3.35 1 15.5 1 0.84 5 19.69 mm
11.16
■
Laterally Loaded Piles
A vertical pile resists a lateral load by mobilizing passive pressure in the soil surrounding it.
(See Figure 11.1c.) The degree of distribution of the soil’s reaction depends on (a) the stiffness
of the pile, (b) the stiffness of the soil, and (c) the fixity of the ends of the pile. In general, laterally loaded piles can be divided into two major categories: (1) short or rigid piles and (2) long
or elastic piles. Figures 11.27a and 11.27b show the nature of the variation of the pile deflection and the distribution of the moment and shear force along the pile length when the pile is
subjected to lateral loading. We next summarize the current solutions for laterally loaded piles.
Elastic Solution
A general method for determining moments and displacements of a vertical pile embedded
in a granular soil and subjected to lateral load and moment at the ground surface was given
by Matlock and Reese (1960). Consider a pile of length L subjected to a lateral force Qg and
592 Chapter 11: Pile Foundations
Deflection
Qg
Shear
Moment
Mg
(a)
Loading Deflection Moment
Qg
Mg
Shear
z
(b)
Figure 11.27 Nature of variation of pile deflection, moment,
and shear force for (a) a rigid pile and (b) and elastic pile
Mg
Qg
x
p
L
z
z
(b)
(a)
x
x
x
x
x
x
p
V
M
z
z
z
z
z
(c)
Figure 11.28 (a) Laterally loaded pile; (b) soil resistance on pile caused by lateral
load; (c) sign conventions for displacement, slope, moment, shear, and soil reaction
11.16 Laterally Loaded Piles
593
a moment Mg at the ground surface (z 5 0), as shown in Figure 11.28a. Figure 11.28b
shows the general deflected shape of the pile and the soil resistance caused by the applied
load and the moment.
According to a simpler Winkler’s model, an elastic medium (soil in this case) can be
replaced by a series of infinitely close independent elastic springs. Based on this assumption,
pr(kN>m)
k5
(11.80)
x(m)
where
k 5 modulus of subgrade reaction
pr 5 pressure on soil
x 5 deflection
The subgrade modulus for granular soils at a depth z is defined as
kz 5 nhz
(11.81)
where nh 5 constant of modulus of horizontal subgrade reaction.
Referring to Figure 11.28b and using the theory of beams on an elastic foundation,
we can write
d4x
EpIp 4 5 pr
(11.82)
dz
where
Ep 5 modulus of elasticity in the pile material
Ip 5 moment of inertia of the pile section
Based on Winkler’s model
pr 5 2kx
(11.83)
The sign in Eq. (11.83) is negative because the soil reaction is in the direction opposite that
of the pile deflection.
Combining Eqs. (11.82) and (11.83) gives
EpIp
d4x
1 kx 5 0
dz4
(11.84)
The solution of Eq. (11.84) results in the following expressions:
Pile Deflection at Any Depth [xz(z)]
xz (z) 5 A x
QgT3
EpIp
1 Bx
MgT2
EpIp
(11.85)
Slope of Pile at Any Depth [uz(z)]
uz (z) 5 A u
QgT2
EpIp
1 Bu
MgT
EpIp
(11.86)
594 Chapter 11: Pile Foundations
Moment of Pile at Any Depth [Mz(z)]
Mz (z) 5 A mQgT 1 BmMg
(11.87)
Shear Force on Pile at Any Depth [Vz(z)]
Vz (z) 5 A vQg 1 Bv
Mg
T
(11.88)
Soil Reaction at Any Depth [ p z9 (z)]
pzr (z) 5 A pr
Qg
T
1 Bpr
Mg
T2
(11.89)
where
A x , Bx , A u , Bu , A m , Bm , A v , Bv , A pr , and Bpr are coefficients
T 5 characteristic length of the soil–pile system
5
EpIp
Å nh
5
(11.90)
nh has been defined in Eq. (11.81)
When L $ 5T, the pile is considered to be a long pile. For L # 2T, the pile is considered to be a rigid pile. Table 11.14 gives the values of the coefficients for long piles
(L>T $ 5) in Eqs. (11.85) through (11.89). Note that, in the first column of the table,
Z5
z
T
(11.91)
is the nondimensional depth.
The positive sign conventions for xz (z), uz (z), Mz (z), Vz (z), and pzr (z) assumed
in the derivations in Table 11.14 are shown in Figure 11.28c. Figure 11.29 shows the variation of A x , Bx , A m , and Bm for various values of L>T 5 Zmax . It indicates that, when
L>T is greater than about 5, the coefficients do not change, which is true of long piles only.
Calculating the characteristic length T for the pile requires assuming a proper value
of nh . Table 11.15 gives some representative values.
Elastic solutions similar to those given in Eqs. 11.85 through 11.89 for piles embedded in cohesive soil were developed by Davisson and Gill (1963). Their equations are
xz (z) 5 A xr
QgR3
EpIp
1 Bxr
MgR2
EpIp
(11.92)
11.16 Laterally Loaded Piles
595
Table 11.14 Coefficients for Long Piles, kz 5 nhz
Z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.2
1.4
1.6
1.8
2.0
3.0
4.0
5.0
Ax
2.435
2.273
2.112
1.952
1.796
1.644
1.496
1.353
1.216
1.086
0.962
0.738
0.544
0.381
0.247
0.142
20.075
20.050
20.009
Au
21.623
21.618
21.603
21.578
21.545
21.503
21.454
21.397
21.335
21.268
21.197
21.047
20.893
20.741
20.596
20.464
20.040
0.052
0.025
Am
0.000
0.100
0.198
0.291
0.379
0.459
0.532
0.595
0.649
0.693
0.727
0.767
0.772
0.746
0.696
0.628
0.225
0.000
20.033
Av
1.000
0.989
0.956
0.906
0.840
0.764
0.677
0.585
0.489
0.392
0.295
0.109
20.056
20.193
20.298
20.371
20.349
20.106
0.015
Ap9
Bx
Bu
0.000
20.227
20.422
20.586
20.718
20.822
20.897
20.947
20.973
20.977
20.962
20.885
20.761
20.609
20.445
20.283
0.226
0.201
0.046
1.623
1.453
1.293
1.143
1.003
0.873
0.752
0.642
0.540
0.448
0.364
0.223
0.112
0.029
20.030
20.070
20.089
20.028
0.000
21.750
21.650
21.550
21.450
21.351
21.253
21.156
21.061
20.968
20.878
20.792
20.629
20.482
20.354
20.245
20.155
0.057
0.049
20.011
Bm
1.000
1.000
0.999
0.994
0.987
0.976
0.960
0.939
0.914
0.885
0.852
0.775
0.688
0.594
0.498
0.404
0.059
20.042
20.026
Bv
0.000
20.007
20.028
20.058
20.095
20.137
20.181
20.226
20.270
20.312
20.350
20.414
20.456
20.477
20.476
20.456
20.213
0.017
0.029
Bp9
0.000
20.145
20.259
20.343
20.401
20.436
20.451
20.449
20.432
20.403
20.364
20.268
20.157
20.047
0.054
0.140
0.268
0.112
20.002
From Drilled Pier Foundations, by R. J. Woodward, W. S. Gardner, and D. M. Greer. Copyright 1972 McGraw-Hill. Used
with permission of the McGraw-Hill Book Company.
Table 11.15 Representative Values of nh
Soil
Dry or moist sand
Loose
Medium
Dense
Submerged sand
Loose
Medium
Dense
nh
kN>m3
1800–2200
5500–7000
15,000–18,000
1000–1400
3500–4500
9000–12,000
and
Mz (z) 5 A m
r QgR 1 Bm
r Mg
(11.93)
where A xr , Bx , A m
r , and Bm
r are coefficients.
and
R5
EpIp
É k
4
(11.94)
596 Chapter 11: Pile Foundations
–1
0
Ax
2
1
3
4
5
0
4
3
Zmax 2
Z z/T
1
2
3
4
4
10
(a)
5
Bx
–1
0
1
2
3
4
0
4
3
Zmax 2
1
Z z/T
2
3
Figure 11.29 Variation of A x , Bx , A m , and Bm
with Z (From Matlock, H. and Reese, L. C.
(1960). “Generalized Solution for Laterally
Loaded Piles,” Journal of the Soil Mechanics and
Foundations Division, American Society of Civil
Engineers, Vol. 86, No. SM5, Part I, pp. 63–91.
With permission from ASCE.)
4, 10
4
5
(b)
The values of the coefficients Ar and Br are given in Figure 11.30 Note that
Z5
z
R
(11.95)
and
Zmax 5
L
R
(11.96)
11.16 Laterally Loaded Piles
0
0.2
Am
0.6
0.4
0.8
1.0
597
1.2
0
Z z/T
1
Zmax 2
2
3
4
3
10
4
5
(c)
Bm
0
0.2
0.4
0.6
0.8
1.0
0
1
Zmax 2
Z z/T
2
3
4
3
10
4
5
(d)
Figure 11.29 (continued)
The use of Eqs. (11.92) and (11.93) requires knowing the magnitude of the characteristic length, R. This can be calculated from Eq. (11.94), provided that the coefficient of
the subgrade reaction is known. For sands, the coefficient of the subgrade reaction was
given by Eq. (11.81), which showed a linear variation with depth. However, in cohesive
598 Chapter 11: Pile Foundations
Ax, Am
–1
0
1
2
0
2
3
Zmax 5
1
2 Zmax
2
Z
3
4
5
3
4
Ax
Am
5
–2
(a)
Bx, Bm
0
–1
0
1
2
4, 5
Zmax 2
1
Zmax 2
3
Z
2
4
3
3
5
r , and
Figure 11.30 Variation of A x , Bxr , A m
Bm
r with Z (From Davisson, M. T. and
Gill, H. L. (1963). “Laterally Loaded Piles in
a Layered Soil System,” Journal of the Soil
Mechanics and Foundations Division,
American Society of Civil Engineers,
Vol. 89, No. SM3, pp. 63–94. With permission
from ASCE.)
4
Bx
Bm
5
(b)
11.16 Laterally Loaded Piles
599
soils, the subgrade reaction may be assumed to be approximately constant with depth.
Vesic (1961) proposed the following equation to estimate the value of k:
EsD4 Es
k 5 0.65 12
Å EpIp 1 2 m2s
(11.97)
Here,
Es 5 modulus of elasticity of soil
D 5 pile width (or diameter)
ms 5 Poisson’s ratio for the soil
For all practical puroses, Eq. (11.97) can be written as
k<
Es
(11.98)
1 2 m2s
Ultimate Load Analysis: Broms’s Method
For laterally loaded piles, Broms (1965) developed a simplified solution based on the
assumptions of (a) shear failure in soil, which is the case for short piles, and (b) bending of the pile, governed by the plastic yield resistance of the pile section, which is
applicable to long piles. Broms’s solution for calculating the ultimate load resistance,
Qu(g) , for short piles is given in Figure 11.31a. A similar solution for piles embedded
in cohesive soil is shown in Figure 11.31b. In Figure 11.31a, note that
Qu(g)
L
e
D
L
0.4 0.2
160
120
Free-headed
pile
8
0.
80
5
1.0 1.
0
.
2
3.0
Restrained
pile
40
4
2
Restrained
pile
cuD2
Ultimate lateral resistance,
Qu(g)
e
L 0
50
0.6
Qu(g)
KpD3
200
Ultimate lateral resistance,
e
D 0
1
60
Free-headed pile
Restrained pile
40
Free-headed
pile
8
Qu(g)
30
16
20
10
0
0
0
4
8
12
L
Length,
D
(a)
16
20
0
4
8
12
16
20
L
Embedment length,
D
(b)
Figure 11.31 Broms’s solution for ultimate lateral resistance of short piles (a) in sand and (b) in clay
600 Chapter 11: Pile Foundations
Kp 5 Rankine passive earth pressure coefficient 5 tan2 ¢ 45 1
fr
≤
2
(11.99)
Similarly, in Figure 11.31b,
cu 5 undrained cohesion <
0.75qu
0.75qu
5
5 0.375qu
FS
2
(11.100)
where
FS 5 factor of safety(52)
qu 5 unconfined compression strength
Figure 11.32 shows Broms’s analysis of long piles. In the figure, the yield moment
for the pile is
My 5 SFY
(11.101)
where
S 5 section modulus of the pile section
FY 5 yield stress of the pile material
In solving a given problem, both cases (i.e., Figure 11.31 and Figure 11.32) should be
checked.
The deflection of the pile head, xz (z 5 0), under working load conditions can be
estimated from Figure 11.33. In Figure 11.33a, the term h can be expressed as
h5
nh
5
Å EpIp
(11.102)
The range of nh for granular soil is given in Table 11.15. Similarly, in Figure 11.33b, which is
for clay, the term K is the horizontal soil modulus and can be defined as
K5
pressure (kN>m2 )
displacement (m)
(11.103)
KD
Å 4EpIp
(11.104)
Also, the term b can be defined as
b5
4
Note that, in Figure 11.33, Qg is the working load.
The following is a general range of values of K for clay soils.
Unconfined compression
strength, qu
kN>m2
200
200–800
. 800
K
kN>m3
10,000–20,000
20,000–40,000
. 40,000
11.16 Laterally Loaded Piles
Free-headed pile
100
Restrained pile
10
e
0
D
1
2
4
8
16
32
Qu(g)
Ultimate lateral resistance,
KpD3
1000
1
0.10
1.0
10.0
Yield moment,
(a)
100.0
My
1000.0
10,000.0
D4Kp
100
60
Restrained
pile
cuD2
20
e
0
D
10
6
Free-headed
pile
8
4
1
4
2
Ultimate lateral resistance,
Qu(g)
40
16
2
1
3
4
6
10
20
40
60
Yield moment,
(b)
100
200
400 600
My
cuD3
Figure 11.32 Broms’s solution for ultimate lateral resistance of long piles
(a) in sand (b) in clay
601
602 Chapter 11: Pile Foundations
Dimensionless lateral deflection,
xz(z 0)(Ep Ip )3/5(nh )2/5
Qg L
10
8
6
Restrained
pile
Free-headed
pile
e
2.0
L
4
1.5
2
1.0
0.8
0.6
0.4
0.2
0.0
0
0
2
4
6
8
10
Dimensionless length, L
(a)
10
e
0.4
L
Dimensionless lateral deflection,
0.2
0.1
xz(z 0)KDL
Qg
8
6
5
0.0
0.0
4
Free-headed
pile
Restrained
pile
2
0
0
1
2
3
4
Dimensionless length, L
(b)
Figure 11.33 Broms’s solution for estimating deflection of pile head
(a) in sand and (b) in clay
5
11.16 Laterally Loaded Piles
603
Example 11.11
Consider a steel H-pile (HP 250 3 85) 25 m long, embedded fully in a granular soil.
Assume that nh 5 12,000 kN>m3 . The allowable displacement at the top of the pile is
8 mm. Determine the allowable lateral load, Qg . Let Mg 5 0. Use the elastic solution.
Solution
From Table 11.1a, for an HP 250 3 85 pile,
Ip 5 123 3 1026 m4
(about the strong axis)
and let
Ep 5 207 3 106 kN>m2
From Eq. (11.90),
T5
EpIp
Å nh
5
5
(207 3 106 ) (123 3 1026 )
5 1.16 m
Å
12,000
5
Here, L>T 5 25>1.16 5 21.55 . 5, so the pile is a long one. Because Mg 5 0,
Eq. (11.85) takes the form
xz (z) 5 A x
QgT3
EpIp
and it follows that
Qg 5
xz (z)EpIp
A xT3
At z 5 0, xz 5 8 mm 5 0.008 m and A x 5 2.435 (see Table 11.14), so
Qg 5
(0.008) (207 3 106 ) (123 3 1026 )
(2.435) (1.163 )
5 53.59 kN
This magnitude of Qg is based on the limiting displacement condition only. However, the magnitude of Qg based on the moment capacity of the pile also needs to be
determined. For Mg 5 0, Eq. (11.87) becomes
Mz (z) 5 A mQgT
According to Table 11.14, the maximum value of A m at any depth is 0.772. The
maximum allowable moment that the pile can carry is
Mz(max) 5 FY
Ip
d1
2
604 Chapter 11: Pile Foundations
Let FY 5 248,000 kN>m2 . From Table 11.1a, Ip 5 123 3 1026 m4 and d1 5 0.254 m,
so
Ip
d1
¢ ≤
2
5
123 3 1026
0.254
¢
≤
2
5 968.5 3 1026 m3
Now,
Qg 5
Mz(max)
A mT
5
(968.5 3 1026 ) (248,000)
5 268.2 kN
(0.772) (1.16)
Because Qg 5 268.2 kN . 53.59 kN,
Qg 5 53.59 kN.
the
deflection
criteria
apply.
Hence,
■
Example 11.12
Solve Example 11.11 by Broms’s method. Assume that the pile is flexible and is free
headed. Let the yield stress of the pile material, Fy 5 248 MN>m2 ; the unit weight of
soil, g 5 18 kN>m3 ; and the soil friction angle fr 5 35°.
Solution
We check for bending failure. From Eq. (11.101),
My 5 SFy
From Table 11.1a,
S5
Ip
d1
2
5
123 3 1026
0.254
2
Also,
My 5
123 3 1026
(248 3 103 ) 5 240.2 kN-m
£ 0.254 §
2
and
My
4
D gKp
5
My
fr
D g tan ¢ 45 1 ≤
2
4
2
5
240.2
35
(0.254) (18) tan ¢ 45 1 ≤
2
4
5 868.8
2
From Figure 11.32a, for My>D4gKp 5 868.8, the magnitude of Qu(g)>KpD3g (for a
free-headed pile with e>D 5 0) is about 140, so
Qu(g) 5 140KpD3g 5 140 tan2 ¢45 1
35
≤ (0.254) 3 (18) 5 152.4 kN
2
11.16 Laterally Loaded Piles
605
Next, we check for pile head deflection. From Eq. (11.102),
h5
nh
12,000
5 5
5 0.86 m21
Å EpIp Å (207 3 106 ) (123 3 1026 )
5
so
hL 5 (0.86) (25) 5 21.5
From Figure 11.33a, for hL 5 21.5, e>L 5 0 (free-headed pile): thus,
xo (EpIp ) 3>5 (nh ) 2>5
QgL
< 0.15
(by interpolation)
and
Qg 5
5
xo (EpIp ) 3>5 (nh ) 2>5
0.15L
(0.008) 3(207 3 106 ) (123 3 1026 )4 3>5 (12,000) 2>5
(0.15) (25)
Hence, Qg 5 40.2 kN (*152.4 kN).
5 40.2 kN
■
Example 11.13
Assume that the 25-m long pile described in Example 11.11 is a restrained pile and is
embedded in clay soil. Given: cu 5 100 kN>m2 and K 5 5,000 kN>m3. The allowable
lateral displacement at the top of the pile is 10 mm. Determine the allowable lateral load
Qg. Given Mymg 5 0. Use Broms’s method.
Solution
From Example 11.12, My 5 240.2 kN-m. So
My
cuD3
5
240.2
5 146.6
(100) (0.254) 3
For the unrestrained pile, from Figure 11.32b,
Qu(g)
cuD2
< 65
or
Qu(g) 5 (65) (100) (0.254) 2 5 419.3 kN
606 Chapter 11: Pile Foundations
Check Pile-Head Deflection
From Eq. (11.104),
b5
KD
(5000) (0.254)
4
5 4
5 0.334
Å 4EpIp Å (4) (207 3 106 ) (123 3 1026 )
bL 5 (0.334) (25) 5 8.35
From Figure 11.33b for bL 5 8.35, by extrapolation the magnitude of
xz (z 5 0)KDL
Qg
Qg 5
xz (z 5 0)KDL
8
5
a
<8
10
b (5000) (0.254) (25)
1000
5 39.7 kN
8
Hence, Qg 5 39.7 kN(* 419.3 kN) .
11.17
■
Pile-Driving Formulas
To develop the desired load-carrying capacity, a point bearing pile must penetrate the dense
soil layer sufficiently or have sufficient contact with a layer of rock. This requirement cannot
always be satisfied by driving a pile to a predetermined depth, because soil profiles vary. For
that reason, several equations have been developed to calculate the ultimate capacity of a pile
during driving. These dynamic equations are widely used in the field to determine whether a
pile has reached a satisfactory bearing value at the predetermined depth. One of the earliest
such equations—commonly referred to as the Engineering News (EN) Record formula—is
derived from the work—energy theory. That is,
Energy imparted by the hammer per blow 5
(pile resistance)(penetration per hammer blow)
According to the EN formula, the pile resistance is the ultimate load Qu , expressed as
Qu 5
where
WR 5 weight of the ram
h 5 height of fall of the ram
S 5 penetration of pile per hammer blow
C 5 a constant
WRh
S1C
(11.105)
11.17 Pile-Driving Formulas
607
The pile penetration, S, is usually based on the average value obtained from the last
few driving blows. In the equation’s original form, the following values of C were recommended:
For drop hammers,
C 5 25.4 mm if S and h are in mm
For steam hammers,
C 5 2.54 mm if S and h are in mm
Also, a factor of safety FS 5 6 was recommended for estimating the allowable pile capacity. Note that, for single- and double-acting hammers, the term WRh can be replaced
by EHE , where E is the efficiency of the hammer and HE is the rated energy of the
hammer. Thus,
Qu 5
EHE
S1C
(11.106)
The EN formula has been revised several times over the years, and other pile-driving
formulas also have been suggested. Three of the other relationships generally used are tabulated in Table 11.16.
The maximum stress developed on a pile during the driving operation can be estimated from the pile-driving formulas presented in Table 11.16. To illustrate, we use the
modified EN formula:
Qu 5
2
EWRh WR 1 n Wp
S 1 C WR 1 Wp
In this equation, S is the average penetration per hammer blow, which can also be
expressed as
S5
25.4
N
(11.107)
where
S is in mm
N 5 number of hammer blows per 25.4 mm of penetration
Thus,
WR 1 n2Wp
EWRh
Qu 5
(25.4>N) 1 2.54 WR 1 Wp
(11.108)
Different values of N may be assumed for a given hammer and pile, and Qu may be
calculated. The driving stress Qu>A p can then be calculated for each value of N. This
608 Chapter 11: Pile Foundations
Table 11.16 Pile-Driving Formulas
Name
Modified EN formula
Formula
2
EWRh WR 1 n Wp
S 1 C WR 1 Wp
where
E 5 efficiency of hammer
C 5 2.54 mm if the units of S and h are in mm
Wp 5 weight of the pile
n 5 coefficient of restitution between the ram
and the pile cap
Qu 5
Typical values for E
Single- and double-acting hammers
Diesel hammers
Drop hammers
0.7–0.85
0.8–0.9
0.7–0.9
Typical values for n
Cast-iron hammer and concrete
piles (without cap)
Wood cushion on steel piles
Wooden piles
Danish formula (Olson and
Flaate, 1967)
EHE
Qu 5
S1
where
Janbu’s formula (Janbu, 1953)
Qu 5
where
0.4–0.5
0.3–0.4
0.25–0.3
EHEL
É 2A pEp
E 5 efficiency of hammer
HE 5 rated hammer energy
Ep 5 modulus of elasticity of the pile material
L 5 length of the pile
A p 5 cross-sectional area of the pile
EHE
Kur S
Kur 5 Cd ¢1 1
Å
11
Cd 5 0.75 1 0.14 ¢
lr 5 ¢
EHEL
A pEpS2
≤
lr
≤
Cd
Wp
WR
≤
11.17 Pile-Driving Formulas
609
procedure can be demonstrated with a set of numerical values. Suppose that a prestressed
concrete pile 24.4 m in length has to be driven by a hammer. The pile sides measure
254 mm. From Table 11.3a, for this pile,
A p 5 645 3 1024 m2
The weight of the pile is
A pLgc 5 (645 3 1024 ) (24.4 m) (23.58 kN>m3 ) 5 37.1 kN
If the weight of the cap is 2.98 kN, then
Wp 5 37.1 1 2.98 5 40.08 kN
For the hammer, let
Rated energy 5 26.03 kN-m 5 HE 5 WRh
Weight of ram 5 22.24 kN
Assume that the hammer efficiency is 0.85 and that n 5 0.35. Substituting these values
into Eq. (11.108) yields
Qu 5
(0.85) (26.03 3 1000) 22.24 1 (0.35) 2 (40.08)
9639.08
c
d 5
kip
£
§
25.4
22.24 1 40.08
25.4
1 2.54
1 2.54
N
N
Now the following table can be prepared:
N
Qu
(kN)
0
2
4
6
8
10
12
20
0
632
1084
1423
1687
1898
2070
2530
Ap
(m2)
645 3
645 3
645 3
645 3
645 3
645 3
645 3
645 3
10 24
10 24
10 24
10 24
10 24
10 24
10 24
10 24
Qu/Ap
(MN/m2)
0
9.79
16.8
22.06
26.16
29.43
32.12
39.22
Both the number of hammer blows per inch and the stress can be plotted in a
graph, as shown in Figure 11.34. If such a curve is prepared, the number of blows per
inch of pile penetration corresponding to the allowable pile-driving stress can easily be
determined.
Actual driving stresses in wooden piles are limited to about 0.7fu . Similarly, for concrete and steel piles, driving stresses are limited to about 0.6fcr and 0.85fy , respectively.
In most cases, wooden piles are driven with a hammer energy of less than 60 kN-m.
Driving resistances are limited mostly to 4 to 5 blows per inch of pile penetration. For concrete and steel piles, the usual values of N are 6 to 8 and 12 to 14, respectively.
610 Chapter 11: Pile Foundations
40
Qu /Ap (MN/m2)
30
20
10
0
0
4
8
12
16
Number of blows /25.4 mm (N)
20 Figure 11.34 Plot of stress versus
blows>25.4 mm.
Example 11.14
A precast concrete pile 0.305 m 3 0.305 m in cross section is driven by a hammer.
Given
Maximum rated hammer energy 5 40.67 kN-m
Hammer efficiency 5 0.8
Weight of ram 5 33.36 kN
Pile length 5 24.39 m
Coefficient of restitution 5 0.4
Weight of pile cap 5 2.45 kN
Ep 5 20.7 3 106 kN>m2
Number of blows for last 25.4 mm of penetration 5 8
Estimate the allowable pile capacity by the
a. Modified EN formula (use FS 5 6)
b. Danish formula (use FS 5 4)
Solution
Part a
Qu 5
2
EWRh WR 1 n Wp
S 1 C WR 1 Wp
Weight of pile 1 cap 5 (0.305 3 0.305 3 24.39) (23.58 kN>m3 ) 1 2.45
5 55.95 kN
11.18 Pile Capacity For Vibration-Driven Piles
611
Given: WRh 5 40.67 kN-m
Qu 5
Qall 5
(0.8) (40.67 3 1000)
25.4
8
3
1 2.54
33.36 1 (0.4) 2 (55.95)
5 2697 kN
33.36 1 55.95
Qu
2697
5
< 449.5 kN
FS
6
Part b
Qu 5
EHE
EHEL
S1
Ä 2A pEp
Use Ep 5 20.7 3 106 kN>m2
EHEL
(0.8) (40.67) (24.39)
5
5 0.01435 m 5 14.35 mm
Å 2A pEp Å 2(0.305 3 0.305) (20.7 3 106 kN>m2 )
Qu 5
Qall 5
11.18
(0.8) (40.67)
25.4
8 3 1000
1 0.01435
< 1857 kN
1857
< 464 kN
4
■
Pile Capacity For Vibration-Driven Piles
The principles of vibratory pile drivers (Figure 11.7e) were discussed briefly in Section 11.4.
As mentioned there, the driver essentially consists of two counterrotating weights. The
amplitude of the centrifugal driving force generated by a vibratory hammer can be given as
Fc 5 mev2
(11.109)
where
m 5 total eccentric rotating mass
e 5 distance between the center of each rotating mass and the center of rotation
v 5 operating circular frequency
Vibratory hammers typically include an isolated bias weight that can range from 4 to
40 kN. The bias weight is isolated from oscillation by springs, so it acts as a net downward
load helping the driving efficiency by increasing the penetration rate of the pile.
The use of vibratory pile drivers began in the early 1930s. Installing piles with vibratory drivers produces less noise and damage to the pile, compared with impact driving.
However, because of a limited understanding of the relationships between the load, the rate
of penetration, and the bearing capacity of piles, this method has not gained popularity in
the United States.
612 Chapter 11: Pile Foundations
Vibratory pile drivers are patented. Some examples are the Bodine Resonant Driver
(BRD), the Vibro Driver of the McKiernan-Terry Corporation, and the Vibro Driver of the
L. B. Foster Company. Davisson (1970) provided a relationship for estimating the ultimate
pile capacity in granular soil:
In SI units,
Qu (kN) 5
0.746(Hp ) 1 98(vp m>s)
(vp m>s) 1 (SL m>cycle) (fHz)
(11.110)
where
Hp 5 horsepower delivered to the pile
vp 5 final rate of pile penetration
SL 5 loss factor
f 5 frequency, in Hz
The loss factor SL for various types of granular soils is as follows (Bowles,
1996):
Closed-End Pipe Piles
•
•
•
Loose sand: 0.244 3 1023 m>cycle
Medium dense sand: 0.762 3 1023 m>cycle
Dense sand: 2.438 3 1023 m>cycle
H-Piles
•
•
•
Loose sand: 20.213 3 1023 m>cycle
Medium dense sand: 0.762 3 1023 m>cycle
Dense sand: 2.134 3 1023 m>cycle
In 2000, Feng and Deschamps provided the following relationship for the ultimate
capacity of vibrodriven piles in granular soil:
Qu 5
3.6(Fc 1 11WB )
LE
vp
L
1 1 1.8 3 1010 "OCR
c
Here,
Fc 5 centrifugal force
WB 5 bias weight
vp 5 final rate of pile penetration
c 5 speed of light 31.8 3 1010 m>min4
OCR 5 overconsolidation ratio
LE 5 embedded length of pile
L 5 pile length
(11.111)
11.19 Negative Skin Friction
613
Example 11.15
Consider a 20-m-long steel pile driven by a Bodine Resonant Driver (Section HP
310 3 125) in a medium dense sand. If Hp 5 350 horsepower, vp 5 0.0016 m>s, and
f 5 115 Hz, calculate the ultimate pile capacity, Qu .
Solution
From Eq. (11.110),
Qu 5
0.746Hp 1 98vp
vp 1 SL f
For an HP pile in medium dense sand, SL < 0.762 3 1023 m>cycle. So
Qu 5
11.19
(0.746) (350) 1 (98) (0.0016)
0.0016 1 (0.762 3 1023 ) (115)
5 2928 kN
■
Negative Skin Friction
Negative skin friction is a downward drag force exerted on a pile by the soil surrounding
it. Such a force can exist under the following conditions, among others:
1. If a fill of clay soil is placed over a granular soil layer into which a pile is driven, the fill will gradually consolidate. The consolidation process will exert a
downward drag force on the pile (see Figure 11.35a) during the period of consolidation.
2. If a fill of granular soil is placed over a layer of soft clay, as shown in Figure 11.35b, it
will induce the process of consolidation in the clay layer and thus exert a downward
drag on the pile.
3. Lowering of the water table will increase the vertical effective stress on the soil at
any depth, which will induce consolidation settlement in clay. If a pile is located in
the clay layer, it will be subjected to a downward drag force.
Sand H
f
fill
Clay H
f
fill
z
L
L
Sand
L1
Neutral
plane
z
Clay
(a)
Figure 11.35 Negative skin friction
(b)
614 Chapter 11: Pile Foundations
In some cases, the downward drag force may be excessive and cause foundation failure. This section outlines two tentative methods for the calculation of negative skin friction.
Clay Fill over Granular Soil (Figure 11.35a)
Similar to the b method presented in Section 11.12, the negative (downward) skin stress
on the pile is
fn 5 Krsor tan dr
(11.112)
where
Kr 5 earth pressure coefficient 5 Ko 5 1 2 sin fr
sor 5 vertical effective stress at any depth z 5 gfr z
gfr 5 effective unit weight of fill
dr 5 soil–pile friction angle < 0.5–0.7fr
Hence, the total downward drag force on a pile is
Hf
Qn 5 3 (pKrgfr tan dr)z dz 5
pKrgfr H 2f tan dr
2
0
(11.113)
where Hf 5 height of the fill. If the fill is above the water table, the effective unit weight,
gfr , should be replaced by the moist unit weight.
Granular Soil Fill over Clay (Figure 11.35b)
In this case, the evidence indicates that the negative skin stress on the pile may exist from
z 5 0 to z 5 L1 , which is referred to as the neutral depth. (See Vesic, 1977, pp. 25 –26.)
The neutral depth may be given as (Bowles, 1982)
L1 5
(L 2 Hf )
L1
B
L 2 Hf
2
1
gfr Hf
gr
R 2
2gfr Hf
gr
(11.114)
where gfr and gr 5 effective unit weights of the fill and the underlying clay layer,
respectively.
For end-bearing piles, the neutral depth may be assumed to be located at the pile tip
(i.e., L1 5 L 2 Hf).
Once the value of L1 is determined, the downward drag force is obtained in
the following manner: The unit negative skin friction at any depth from z 5 0 to
z 5 L1 is
fn 5 Krsor tan dr
where
Kr 5 Ko 5 1 2 sin fr
sor 5 gfr Hf 1 grz
dr 5 0.5–0.7fr
(11.115)
11.19 Negative Skin Friction
L1
615
L1
Qn 5 3 pfn dz 5 3 pKr(gfr Hf 1 grz)tan dr dz
0
0
5 (pKrgfr Hf tan dr)L1 1
L21pKrgr tan dr
2
(11.116)
If the soil and the fill are above the water table, the effective unit weights should be
replaced by moist unit weights. In some cases, the piles can be coated with bitumen in the
downdrag zone to avoid this problem.
A limited number of case studies of negative skin friction is available in the literature. Bjerrum et al. (1969) reported monitoring the downdrag force on a test pile at
Sorenga in the harbor of Oslo, Norway (noted as pile G in the original paper). The study
of Bjerrum et al. (1969) was also discussed by Wong and Teh (1995) in terms of the pile
being driven to bedrock at 40 m. Figure 11.36a shows the soil profile and the pile. Wong
and Teh estimated the following quantities:
Fill: Moist unit weight, gf 5 16 kN>m3
Saturated unit weight, gsat(f) 5 18.5 kN>m3
So
gfr 5 18.5 2 9.81 5 8.69 kN>m3
and
Hf 5 13 m
•
f 16 kN/m3
2m
11 m
0
0
Fill
Groundwater
table
Axial force in pile (kN)
1000
2000
3000
Fill
sat ( f ) 18.5 kN/m3
40 m
Pile D 500 mm
Depth (m)
10
20
30
Clay
40
Rock
(a)
(b)
Figure 11.36 Negative skin friction on a pile in the harbor of Oslo, Norway
(Based on Bjerrum et al., (1969) and Wong and The (1995))
616 Chapter 11: Pile Foundations
Clay: Kr tan dr < 0.22
Saturated effective unit weight, gr 5 19 2 9.81 5 9.19 kN>m3
Pile: L 5 40 m
Diameter, D 5 500 m
•
•
Thus, the maximum downdrag force on the pile can be estimated from Eq. (11.116).
Since in this case the pile is a point bearing pile, the magnitude of L1 5 27 m, and
Qn 5 (p) (Kr tan dr) 3gf 3 2 1 (13 2 2)gfr 4 (L1 ) 1
L21pgr(Kr tan dr)
2
or
Qn 5 (p 3 0.5)(0.22) 3(16 3 2) 1 (8.69 3 11)4 (27) 1
(27) 2 (p 3 0.5)(9.19)(0.22)
2
5 2348 kN
The measured value of the maximum Qn was about 2500 kN (Figure 11.36b), which is in
good agreement with the calculated value.
Example 11.16
In Figure 11.35a, let Hf 5 2m. The pile is circular in cross section with a diameter of
0.305 m. For the fill that is above the water table, gf 5 16 kN>m3 and fr 5 32°. Determine the total drag force. Use dr 5 0.6fr.
Solution
From Eq. (11.113),
Qn 5
pK rgfH 2f tan dr
2
with
p 5 p(0.305) 5 0.958m
K r 5 1 2 sin fr 5 1 2 sin 32 5 0.47
and
dr 5 (0.6) (32) 5 19.2°
Thus,
Qn 5
(0.958) (0.47) (16) (2) 2 tan 19.2
5 5.02 kN
2
■
Example 11.17
In Figure 11.35b, let Hf 5 2 m, pile diameter 5 0.305 m, gf 5 16.5 kN>m3,
fclay
r 5 34°, gsat(clay) 5 17.2 kN>m3, and L 5 20 m. The water table coincides with the
r .
top of the clay layer. Determine the downward drag force. Assume that dr 5 0.6fclay
11.20 Group Efficiency
617
Solution
The depth of the neutral plane is given in Eq. (11.114) as
L1 5
L 2 Hf
L1
a
L 2 Hf
2
1
gfHf
gr
b2
2gfHf
gr
Note that gfr in Eq. (11.114) has been replaced by gf because the fill is above the water
table, so
L1 5
(20 2 2) (20 2 2)
(16.5) (2)
(2) (16.5) (2)
c
1
d 2
L1
2
17.2 2 9.81)
(17.2 2 9.81)
or
L1 5
242.4
2 8.93; L1 5 11.75 m
L1
Now, from Eq. (11.116), we have
Qn 5 (pK rgfHf tan dr )L1 1
L21pK rgr tan dr
2
with
p 5 p(0.305) 5 0.958m
and
K r 5 1 2 sin 34° 5 0.44
Hence,
Qn 5 (0.958) (0.44) (16.5) (2) 3tan(0.6 3 34)4 (11.75)
1
(11.75) 2 (0.958) (0.44) (17.2 2 9.81) 3tan(0.6 3 34) 4
5 60.78 1 79.97 5 140.75 kN
2
■
Group Piles
11.20
Group Efficiency
In most cases, piles are used in groups, as shown in Figure 11.37, to transmit the structural
load to the soil. A pile cap is constructed over group piles. The cap can be in contact with
the ground, as in most cases (see Figure 11.37a), or well above the ground, as in the case
of offshore platforms (see Figure 11.37b).
Determining the load-bearing capacity of group piles is extremely complicated and
has not yet been fully resolved. When the piles are placed close to each other, a
618 Chapter 11: Pile Foundations
Pile cap
Section
Water table
L
d
d
d
d
L
Plan
(b)
d
Bg
Lg
d
Number of piles in group n1 n2
(Note: Lg Bg)
Lg (n1 1) d 2(D/2)
Bg (n2 1) d 2(D/2)
(a)
(c)
Figure 11.37 Group piles
reasonable assumption is that the stresses transmitted by the piles to the soil will overlap
(see Figure 11.37c), reducing the load-bearing capacity of the piles. Ideally, the piles in
a group should be spaced so that the load-bearing capacity of the group is not less than
the sum of the bearing capacity of the individual piles. In practice, the minimum centerto-center pile spacing, d, is 2.5D and, in ordinary situations, is actually about 3 to 3.5D.
11.20 Group Efficiency
619
The efficiency of the load-bearing capacity of a group pile may be defined as
h5
Qg(u)
S Qu
(11.117)
where
h 5 group efficiency
Qg(u) 5 ultimate load-bearing capacity of the group pile
Qu 5 ultimate load-bearing capacity of each pile without the group effect
Many structural engineers use a simplified analysis to obtain the group efficiency for
friction piles, particularly in sand. This type of analysis can be explained with the aid of
Figure 11.37a. Depending on their spacing within the group, the piles may act in one of two
ways: (1) as a block, with dimensions Lg 3 Bg 3 L, or (2) as individual piles. If the piles
act as a block, the frictional capacity is favpgL < Qg(u) . [Note: pg 5 perimeter of the cross
section of block 5 2(n1 1 n2 2 2)d 1 4D, and fav 5 average unit frictional resistance.]
Similarly, for each pile acting individually, Qu < pLfav . (Note: p 5 perimeter of the cross
section of each pile.) Thus,
h5
5
Qg(u)
S Qu
5
fav 32(n1 1 n2 2 2)d 1 4D4L
n1n2pLfav
2(n1 1 n2 2 2)d 1 4D
pn1n2
(11.118)
Hence,
Qg(u) 5 B
2(n1 1 n2 2 2)d 1 4D
RS Qu
pn1n2
(11.119)
From Eq. (11.119), if the center-to-center spacing d is large enough, h . 1. In that case,
the piles will behave as individual piles. Thus, in practice, if h , 1, then
Qg(u) 5 hS Qu
and if h $ 1, then
Qg(u) 5 S Qu
There are several other equations like Eq. (11.119) for calculating the group efficiency of friction piles. Some of these are given in Table 11.17.
It is important, however, to recognize that relationships such as Eq. (11.119) are simplistic and should not be used. In fact, in a group pile, the magnitude of fav depends on the
location of the pile in the group (ex., Figure 11.38).
620 Chapter 11: Pile Foundations
Table 11.17 Equations for Group Efficiency of Friction Piles
Name
Equation
Converse–Labarre equation
h512 B
(n1 2 1)n2 1 (n2 2 1)n1
90n1n2
Ru
where u(deg) 5 tan21 (D>d)
Los Angeles Group Action
equation
h512
D
3n (n 2 1)
pdn1n2 1 2
1 n2 (n1 2 1) 1 "2(n1 2 1) (n2 2 1) 4
Seiler–Keeney equation
(Seiler and Keeney, 1944)
h 5 b1 2 B
n1 1 n2 2 2
0.3
11d
RB
Rr 1
n1 1 n2 2 1
n1 1 n2
7(d2 2 1)
where d is in ft
Sandy soil
L
= 18
D
D = 250 mm
Average skin friction, f (kN/m2)
50
Center pile
Border pile
40
Corner pile
30
20
10
1.5D
3D
0
10
20
Settlement (mm)
30
Figure 11.38 Average skin friction (fav ) based on pile location
(Based on Liu et al., 1985)
11.21 Ultimate Capacity of Group Piles in Saturated Clay
621
3
d
d
D
Group efficiency, d
30°
2
d
35°
40°
1
45°
0
0
1
2
4
d
D
6
8
Figure 11.39 Variation
of efficiency of pile
groups in sand (Based
on Kishida and
Meyerhof, 1965)
Figure 11.39 shows the variation of the group efficiency h for a 3 3 3 group pile in
sand (Kishida and Meyerhof, 1965). It can be seen that, for loose and medium sands, the
magnitude of the group efficiency can be larger than unity. This is due primarily to the densification of sand surrounding the pile.
11.21
Ultimate Capacity of Group Piles in Saturated Clay
Figure 11.40 shows a group pile in saturated clay. Using the figure, one can estimate the
ultimate load-bearing capacity of group piles in the following manner:
Step 1. Determine S Qu 5 n1n2 (Qp 1 Qs ). From Eq. (11.18),
Qp 5 A p 39cu(p) 4
where cu(p) 5 undrained cohesion of the clay at the pile tip.
Also, from Eq. (11.55),
Qs 5 S apcu DL
So,
S Qu 5 n1n2 39A pcu(p) 1 S apcu DL4
(11.120)
Step 2. Determine the ultimate capacity by assuming that the piles in the group act as
a block with dimensions Lg 3 Bg 3 L. The skin resistance of the block is
S pgcu DL 5 S 2(Lg 1 Bg )cu DL
Calculate the point bearing capacity:
A pqp 5 A pcu(p)N *c 5 (LgBg )cu(p)N *c
622 Chapter 11: Pile Foundations
Qg(u)
2 (Lg Bg)cu L
cu cu(1)
L
cu cu(2)
cu cu(3)
Lg Bg cu (p) N *c
Bg
Figure 11.40 Ultimate capacity of
group piles in clay
Lg
Obtain the value of the bearing capacity factor N c* from Figure 11.41.
Thus, the ultimate load is
S Qu 5 LgBgcu(p)N c* 1 S 2(Lg 1 Bg )cu DL
(11.121)
Step 3. Compare the values obtained from Eqs. (11.120) and (11.121). The lower of
the two values is Qg(u) .
9
Lg/Bg 1
8
2
3
N *c
7
6
5
4
0
1
2
3
L/Bg
Figure 11.41 Variation of N c* with Lg>Bg and L>Bg
4
5
11.21 Ultimate Capacity of Group Piles in Saturated Clay
623
Example 11.18
The section of a 3 3 4 group pile in a layered saturated clay is shown in Figure 11.42. The
piles are square in cross section (356 mm 3 356 mm). The center-to-center spacing, d, of
the piles is 889 mm. Determine the allowable load-bearing capacity of the pile group. Use
FS 5 4. Note that the groundwater table coincides with the ground surface.
Solution
From Eq. (11.120),
SQu 5 n1n2 39A pcu(p) 1 a1pcu(1)L1 1 a2pcu(2)L24
From Figure 11.42, cu(1) 5 50.3 kN>m2 and cu(2) 5 85.1 kN>m2.
For the top layer with cu(1) 5 50.3 kN>m2,
cu(1)
pa
5
50.3
5 0.503
100
From Table 11.10, a1 < 0.68. Similarly,
cu(2)
pa
5
85.1
< 0.85
100
a2 5 0.51
SQu 5 (3) (4) c
(9) (0.356) 2 (85.1) 1 (0.68) (4 3 0.356) (50.3) (4.57)
d
1 (0.51) (4 3 0.356) (85.1) (13.72)
5 14011 kN
For piles acting as a group.
Lg 5 (3) (0.889) 1 0.356 5 3.023 m
Bg 5 (2) (0.889) 1 0.356 5 2.134 m
G.W.T.
Clay
cu 50.3 kN/m2
sat 17.6 kN/m3
4.57 m
Clay
cu 85.1 kN/m2
sat 19.02 kN/m3
13.72 m
889 mm
Figure 11.42 Group pile of
layered saturated clay
624 Chapter 11: Pile Foundations
Lg
Bg
5
3.023
5 1.42
2.134
L
18.29
5
5 8.57
Bg
2.134
From Figure 11.41, N *c 5 8.75. From Eq. (11.121),
SQu 5 LgBgcu(p)N*c 1 S2(Lg 1 Bg )cu DL
5 (3.023) (2.134) (85.1) (8.75) 1 (2) (3.023 12.134) 3(50.3) (4.57)
1 (85.1) (13.72) 4
5 19217 kN
Hence, SQu 5 14,011 kN.
SQall 5
11.22
14,011
14,011
5
< 3503 kN
FS
4
■
Elastic Settlement of Group Piles
In general, the settlement of a group pile under a similar working load per pile increases
with the width of the group (Bg ) and the center-to-center spacing of the piles (d). Several
investigations relating to the settlement of group piles have been reported in the literature,
with widely varying results. The simplest relation for the settlement of group piles was
given by Vesic (1969), namely,
sg(e) 5
Bg
ÉD
se
(11.122)
where
sg(e) 5 elastic settlement of group piles
Bg 5 width of group pile section
D 5 width or diameter of each pile in the group
se 5 elastic settlement of each pile at comparable working load (see Section 11.15)
For group piles in sand and gravel, for elastic settlement, Meyerhof (1976) suggested
the empirical relation
sg(e) (mm) 5
0.96q"BgI
N60
(11.123)
11.22 Elastic Settlement of Group Piles
625
where
q 5 Qg> (LgBg ) (in kN>m2 )
(11.124)
and
Lg and Bg 5 length and width of the group pile section, respectively (m)
N60 5 average standard penetration number within seat of settlement (<Bg deep
below the tip of the piles)
(11.125)
I 5 influence factor 5 1 2 L>8Bg > 0.5
L 5 length of embedment of piles (m)
Similarly, the group pile settlement is related to the cone penetration resistance by the
formula
Sg(e) 5
qBgI
2qc
(11.126)
where qc 5 average cone penetration resistance within the seat of settlement. (Note that,
in Eq. (11.126), all quantities are expressed in consistent units.)
Example 11.19
Consider a 3 3 4 group of prestressed concrete piles, each 21 m long, in a sand layer.
The details of each pile and the sand are similar to that described in Example 11.10. The
working load for the pile group is 6024 kN (3 3 4 3 Qall—where Qall 5 502 kN as in
Example 11.10), and d>D 5 3. Estimate the elastic settlement of the pile group. Use
Eq. (11.123).
Solution
se(g) 5
Bg
ÉD
se
Bg 5 (3 2 1)d 1
2D
5 (2) (3D) 1 D 5 7D 5 (7) (0.356 m) 5 2.492 m
2
From Example 11.10, se 5 19.69 mm. Hence,
se(g) 5
2.492
(19.69) 5 52.09 mm
É 0.356
■
626 Chapter 11: Pile Foundations
11.23
Consolidation Settlement of Group Piles
The consolidation settlement of a group pile in clay can be estimated by using the
2:1 stress distribution method. The calculation involves the following steps (see
Figure 11.43):
Step 1. Let the depth of embedment of the piles be L. The group is subjected
to a total load of Qg . If the pile cap is below the original ground surface,
Qg equals the total load of the superstructure on the piles, minus
the effective weight of soil above the group piles removed by
excavation.
Step 2. Assume that the load Qg is transmitted to the soil beginning at a depth of
2L>3 from the top of the pile, as shown in the figure. The load Qg spreads
out along two vertical to one horizontal line from this depth. Lines aar and
bbr are the two 2:1 lines.
Step 3. Calculate the increase in effective stress caused at the middle of each soil
layer by the load Qg . The formula is
Dsir 5
Qg
(Bg 1 zi ) (Lg 1 zi )
Bg
Lg
Qg
Clay layer 1
Groundwater
table
L
a
b
2
L
3
Clay layer 2
L1
z
2V:1H
Clay layer 3
L2
2V:1H
Clay layer 4
a
L3
b
Rock
Figure 11.43 Consolidation settlement of group piles
(11.127)
11.23 Consolidation Settlement of Group Piles
627
where
Dsir 5 increase in effective stress at the middle of layer i
Lg , Bg 5 length and width, respectively of the planned group piles
zi 5 distance from z 5 0 to the middle of the clay layer i
For example, in Figure 11.43, for layer 2, zi 5 L1>2; for layer 3, zi 5
L1 1 L2>2; and for layer 4, zi 5 L1 1 L2 1 L3>2. Note, however, that
there will be no increase in stress in clay layer 1, because it is above the horizontal plane (z 5 0) from which the stress distribution to the soil starts.
Step 4. Calculate the consolidation settlement of each layer caused by the increased
stress. The formula is
Dsc(i) 5 B
De(i)
1 1 eo(i)
RHi
(11.128)
where
Dsc(i) 5 consolidation settlement of layer i
De(i) 5 change of void ratio caused by the increase in stress in layer i
eo(i) 5 initial void ratio of layer i (before construction)
Hi 5 thickness of layer i (Note: In Figure 11.43, for layer 2, Hi 5 L1 ;
for layer 3, Hi 5 L2 ; and for layer 4, Hi 5 L3 .)
Relationships involving De(i) are given in Chapter 1.
Step 5. The total consolidation settlement of the group piles is then
Dsc(g) 5 SDsc(i)
(11.129)
Note that the consolidation settlement of piles may be initiated by fills placed nearby,
adjacent floor loads, or the lowering of water tables.
Example 11.20
A group pile in clay is shown in Figure 11.44. Determine the consolidation settlement
of the piles. All clays are normally consolidated.
Solution
Because the lengths of the piles are 15 m each, the stress distribution starts at a depth of
10 m below the top of the pile. We are given that Qg 5 2000 kN.
Calculation of Settlement of Clay Layer 1
For normally consolidated clays,
Dsc(1) 5 c
Ds(1)
r 5
(Cc(1)H1 )
1 1 eo(1)
d log c
so(1)
r 1 Ds(1)
r
Qg
(Lg 1 z1 ) (Bg 1 z1 )
so(1)
r
5
d
2000
5 51.6 kN>m2
(3.3 1 3.5) (2.2 1 3.5)
628 Chapter 11: Pile Foundations
Qg 2000 kN
Sand
16.2 kN/m3
Groundwater
table
10 m
15 m
2m
1m
Clay
9m
Group
pile
16 m
7m
z
2V:1H
o(1)
(1)
Clay
o(2)
(2)
4m
o(3), (3)
2m
Clay
2V:1H
Rock
sat 18.0 kN/m3
eo 0.82
Cc 0.3
sat 18.9 kN/m3
eo 0.7
Cc 0.2
sat 19 kN/m3
eo 0.75
Cc 0.25
Pile group: Lg 3.3 m; Bg 2.2 m
(not to scale)
Figure 11.44 Consolidation settlement of a pile group
and
so(1)
r 5 2(16.2) 1 12.5(18.0 2 9.81) 5 134.8 kN>m2
So
Dsc(1) 5
(0.3) (7)
134.8 1 51.6
log c
d 5 0.1624 m 5 162.4 mm
1 1 0.82
134.8
Settlement of Layer 2
As with layer 1,
Dsc(2) 5
Cc(2)H2
1 1 eo(2)
log c
so(2)
r 1 Ds(2)
so(2)
r
d
ss(2)
r
5 2(16.2) 1 16(18.0 2 9.81) 1 2(18.9 2 9.81) 5 181.62 kN>m2
and
Ds(2)
r 5
2000
5 14.52 kN>m2
(3.3 1 9) (2.2 1 9)
Hence,
Dsc(2) 5
(0.2) (4)
181.62 1 14.52
log c
d 5 0.0157 m 5 15.7 mm
1 1 0.7
181.62
Problems
629
Settlement of Layer 3
Continuing analogously, we have
so(3)
r 5 181.62 1 2(18.9 2 9.81) 1 1(19 2 9.81) 5 208.99 kN>m2
Ds(3)
r 5
2000
5 9.2 kN>m2
(3.3 1 12) (2.2 1 12)
Dsc(3) 5
(0.25) (2)
208.99 1 9.2
log a
b 5 0.0054 m 5 5.4 mm
1 1 0.75
208.99
Hence, the total settlement is
Dsc(g) 5 162.4 1 15.7 1 5.4 5 183.5 mm
11.24
■
Piles in Rock
For point bearing piles resting on rock, most building codes specify that Qg(u) 5 S Qu ,
provided that the minimum center-to-center spacing of the piles is D 1 300 mm. For Hpiles and piles with square cross sections, the magnitude of D is equal to the diagonal
dimension of the cross section of the pile.
Problems
11.1 A 12 m long concrete pile is shown in Figure P11.1. Estimate the ultimate point
load Qp by
a. Meyerhof’s method
b. Vesic’s method
c. Coyle and Castello’s method
Use m 5 600 in Eq. (11.26).
11.2 Refer to the pile shown in Figure P11.1. Estimate the side resistance Qs by
a. Using Eqs. (11.40) through (11.42). Use K 5 1.3 and dr 5 0.8fr
b. Coyle and Castello’s method [Eq. (11.44)]
11.3 Based on the results of Problems 11.1 and 11.2, recommend an allowable load for
the pile. Use FS 5 4.
11.4 A driven closed-ended pile, circular in cross section, is shown in Figure P11.4.
Calculate the following.
a. The ultimate point load using Meyerhof’s procedure.
b. The ultimate point load using Vesic’s procedure. Take Irr 5 50.
c. An approximate ultimate point load on the basis of parts (a) and (b).
d. The ultimate frictional resistance Qs. [Use Eqs. (11.40) through (11.42), and
take K 5 1.4 and dr 5 0.6fr.]
e. The allowable load of the pile (use FS 5 4).
11.5 Following is the variation of N60 with depth in a granular soil deposit. A concrete
pile 9 m long (0.305 m 3 0.305 m in cross section) is driven into the sand and
fully embedded in the sand.
630 Chapter 11: Pile Foundations
Concrete pile
356 mm 356 mm
Loose sand
1 30º
17.5 kN/m3
12 m
Dense sand
2 42º
18.5 kN/m3
3.05 m
Groundwater
table
3.05 m
Figure P11.1
15.72 kN/m3
32º
c 0
sat 18.24 kN/m3
32°
c 0
sat 19.24 kN/m3
40º
c 0
15.24 m
Figure P11.4
15 in.
Depth (m)
N60
1.5
3.0
4.5
6.0
7.5
9.0
10.5
11.0
12.5
14.0
4
8
7
5
16
18
21
24
20
19
Problems
631
Silty clay
sat 18.55 kN/m3
cu 35 kN/m2
6.1 m
Groundwater
table
Silty clay
sat 19.24 kN/m3
cu 75 kN/m2
12.2 m
406 mm
11.6
11.7
11.8
11.9
11.10
11.11
11.12
Figure P11.10
Estimate the allowable load-carrying capacity of the pile (Qall). Use FS 5 4 and
Meyerhof’s equations [Eqs. (11.37) and (11.45)].
Solve Problem 11.5 using the equation of Briaud et al. [Eqs. (11.38) and (11.47)].
A concrete pile 15.24 m long having a cross section of 406 mm 3 406 mm is
fully embedded in a saturated clay layer for which gsat 5 19.02 kN>m3, f 5 0,
and cu 5 76.7 kN>m2. Determine the allowable load that the pile can carry. (Let FS
5 3.) Use the a method to estimate the skin friction and Veric’s method for point
load estimation.
Redo Problem 11.7 using the l method for estimating the skin friction and
Meyerhof’s method for the point load estimation.
A concrete pile 15 m long having a cross section of 0.38 m 3 0.38 m is fully embedded in a saturated clay layer. For the clay, given: gsat 5 18 kN>m3, f 5 0, and
cu 5 80 kN>m2. Determine the allowable load that the pile can carry (FS 5 3). Use
the l method to estimate the skin resistance.
A concrete pile 406 mm 3 406 mm in cross section is shown in Figure P11.10.
Calculate the ultimate skin friction resistance by using the
a. a method
b. l method
c. b method
Use fRr 5 20° for all clays, which are normally consolidated.
A steel pile (H-section; HP 360 3 152; see Table 11.1) is driven into a layer of
sandstone. The length of the pile is 18.9 m. Following are the properties of the
sandstone: unconfined compression strength 5 qu(lab) 5 78.7 MN/m2 and angle of
friction 5 36°. Using a factor of safety of 3, estimate the allowable point load that
can be carried by the pile. Use 3qu(design) 5 qu(lab)>54.
A concrete pile is 18 m long and has a cross section of 0.406 m 3 0.406 m. The
pile is embedded in a sand having g 5 16 kN>m3 and fr 5 37°. The allowable
632 Chapter 11: Pile Foundations
11.13
11.14
11.15
11.16
11.17
11.18
11.19
11.20
11.21
d
working load is 900 kN. If 600 kN are contributed by the frictional resistance and
300 kN are from the point load, determine the elastic settlement of the pile. Given:
Ep 5 2.1 3 106 kN/m.2, Es 5 30 3 103 kN/m.2, ms 5 0.38, and j 5 0.57
[Eq. (11.73)].
Solve Problem 11.12 with the following: length of pile 5 15 m, pile cross section 5
0.305 m 3 0.305 m, allowable working load 5 338 kN, contribution of frictional resistance to working load 5 280 kN, Ep 5 21 3 106 kN>m2, Es 5 30,000 kN>m2,
ms 5 0.3, and j 5 0.62 [Eq. (11.73)].
A 30-m long concrete pile is 305 mm 3 305 mm in cross section and is fully embedded in a sand deposit. If nh 5 9200 kN>m2, the moment at ground level,
Mg 5 0, the allowable displacement of pile head 5 12 mm; Ep 5 21 3 106 kN>m2;
and FY (pile) 5 21,000 kN>m2, calculate the allowable lateral load, Qg, at the ground
level. Use the elastic solution method.
Solve Problem 11.14 by Brom’s method. Assume that the pile is flexible and free
headed. Let the soil unit weight, g 5 16 kN>m3; the soil friction angle, fr 5 30°;
and the yield stress of the pile material, FY 5 21 MN>m2.
A steel H-pile (section HP 330 3 149) is driven by a hammer. The maximum
rated hammer energy is 54.23 kN-m, the weight of the ram is 53.4 kN, and the
length of the pile is 27.44 m. Also, we have coefficient of restitution 5 0.35,
weight of the pile cap 5 10.7 kN, hammer efficiency 5 0.85, number of blows
for the last inch of penetration 5 10, and Ep 5 207 3 106 kN>m2. Estimate the
pile capacity using Eq. (11.106). Take FS 5 6.
Solve Problem 11.16 using the modified EN formula. (See Table 11.16). Use
FS 5 4.
Solve Problem 11.16 using the Danish formula (See Table 11.16). Use FS 5 3.
Figure 11.35a shows a pile. Let L 5 20 m, D (pile diameter) 5 450 mm,
Hf 5 4 m, gfill 5 17.5 kN>m3, and frfill 5 25°. Determine the total downward
drag force on the pile. Assume that the fill is located above the water table and
r .
that d r 5 0.5ffill
Redo Problem 11.19 assuming that the water table coincides with the top of the fill
and that gsat(fill) 5 19.8 kN>m3 . If the other quantities remain the same, what
would be the downward drag force on the pile? Assume dr 5 0.5ffill
r .
3
Refer to Figure 11.35b. Let L 5 15.24 m, gfill 5 17.29 kN>m , gsat(clay) 5
19.49 kN>m3, fclay
r 5 20°, Hf 5 3.05 m, and D (pile diameter) 5 406 mm. The
water table coincides with the top of the clay layer. Determine the total downward
drag force on the pile. Assume dr 5 0.6fclay
r .
Figure p11.23
Problems
633
Clay
cu 25 kN/m2
5m
Clay
cu 45 kN/m2
6m
6m
Clay
cu 60 kN/m2
1m
Figure P11.25
11.22 A concrete pile measuring 0.406 m 3 0.406 m in cross section is 18.3 m long. It is
fully embedded in a layer of sand. The following is an approximation of the mechanical cone penetration resistance (qc ) and the friction ratio (Fr ) for the sand
layer. Estimate the allowable bearing capacity of the pile. Use FS 5 4.
Depth below ground surface (m)
0–6.1
6.1–13.7
13.7–19.8
qc (kN , m2 )
Fr (%)
2803
3747
8055
2.3
2.7
2.8
11.23 The plan of a group pile is shown in Figure P11.23. Assume that the piles are
embedded in a saturated homogeneous clay having a cu 5 86 kN>m2. Given:
diameter of piles (D) 5 316 mm, center-to-center spacing of piles 5 600 mm,
and length of piles 5 20 m. Find the allowable load-carrying capacity of the pile
group. Use FS 5 3.
11.24 Redo Problem 11.23 with the following: center-to-center spacing of piles 5 762 mm,
length of piles 5 13.7 m, D 5 305 mm, cu 5 41.2 kN>m2, gsat 5 19.24 kN>m3,
and FS 5 3.
11.25 The section of a 4 3 4 group pile in a layered saturated clay is shown in Figure P11.25.
The piles are square in cross section (356 mm 3 356 mm). The center-to-center
spacing (d) of the piles is 1 m. Determine the allowable load-bearing capacity of the
pile group. Use FS 5 3.
11.26 Figure P11.26 shows a group pile in clay. Determine the consolidation settlement
of the group. Use the 2:1 method of estimate the average effective stress in the
clay layers.
634 Chapter 11: Pile Foundations
1335 kN
3m
Groundwater
table
Sand
= 15.72 kN/m3
Sand
sat = 18.55 kN/m3
3m
2.75 m
2.75 m
Group
plan
18 m
Normally consolidated clay
sat = 19.18 kN/m3
eo = 0.8
Cc = 0.8
5m
Normally consolidated clay
sat = 18.08 kN/m3
eo = 1.0
Cc = 0.31
3m
Normally consolidated clay
sat = 19.5 kN/m3
eo = 0.7
Cc = 0.26
15 m
Rock
Figure P11.26
References
AMERICAN SOCIETY OF CIVIL ENGINEERS (1959). “Timber Piles and Construction Timbers,” Manual
of Practice, No. 17, American Society of Civil Engineers, New York.
AMERICAN SOCIETY OF CIVIL ENGINEERS (1993). Design of Pile Foundations (Technical Engineering
and Design Guides as Adapted from the U.S. Army Corps of Engineers, No. 1), American
Society of Civil Engineers, New York.
BALDI, G., BELLOTTI, R., GHIONNA, V., JAMIOLKOWSKI, M. and PASQUALINI, E. (1981). “Cone Resistance in Dry N.C. and O.C. Sands, Cone Penetration Testing and Experience,” Proceedings,
ASCE Specialty Conference, St. Louis, pp. 145–177.
BJERRUM, L., JOHANNESSEN, I. J., and EIDE, O. (1969). “Reduction of Skin Friction on Steel Piles to
Rock,” Proceedings, Seventh International Conference on Soil Mechanics and Foundation
Engineering, Mexico City, Vol. 2, pp. 27–34.
References
635
BOWLES, J. E. (1982). Foundation Analysis and Design, McGraw-Hill, New York.
BOWLES, J. E. (1996). Foundation Analysis and Design, McGraw-Hill, New York.
BRIAUD, J. L., TUCKER, L., LYTTON, R. L., and COYLE, H. M. (1985). Behavior of Piles and Pile Groups,
Report No. FHWA>RD-83>038, Federal Highway Administration, Washington, DC.
BROMS, B. B. (1965). “Design of Laterally Loaded Piles,” Journal of the Soil Mechanics and
Foundations Division, American Society of Civil Engineers, Vol. 91, No. SM3, pp. 79 – 99.
CHEN, Y. J., and KULHAWY, F. H. (1994). “Case History Evaluation of the Behavior of Drilled Shafts
under Axial and Lateral Loading,” Final Report, Project 1493-04, EPRI TR-104601, Geotechnical Group, Cornell University, Ithaca, NY, December.
COYLE, H. M., and CASTELLO, R. R. (1981). “New Design Correlations for Piles in Sand,” Journal of
the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 107,
No. GT7, pp. 965–986.
DAVISSON, M. T. (1973). “High Capacity Piles” in Innovations in Foundation Construction, Proceedings of a Lecture Series, Illinois Section, American Society of Civil Engineers, Chicago.
DAVISSON, M. T. (1970). “BRD Vibratory Driving Formula,” Foundation Facts, Vol. VI, No. 1, pp. 9–11.
DAVISSON, M. T., and GILL, H. L. (1963). “Laterally Loaded Piles in a Layered Soil System,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers,
Vol. 89, No. SM3, pp. 63 –94.
FENG, Z., and DESCHAMPS, R. J. (2000). “A Study of the Factors Influencing the Penetration and Capacity of Vibratory Driven Piles,” Soils and Foundations, Vol. 40, No. 3, pp. 43–54.
GOODMAN, R. E. (1980). Introduction to Rock Mechanics, Wiley, New York.
GUANG-YU, Z. (1988). “Wave Equation Applications for Piles in Soft Ground,” Proceedings, Third
International Conference on the Application of Stress-Wave Theory to Piles (B. H. Fellenius,
ed.), Ottawa, Ontario, Canada, pp. 831–836.
JANBU, N. (1953). An Energy Analysis of Pile Driving with the Use of Dimensionless Parameters,
Norwegian Geotechnical Institute, Oslo, Publication No. 3.
KISHIDA, H., and MEYERHOF, G. G. (1965). “Bearing Capacity of Pile Groups under Eccentric Loads
in Sand,” Proceedings, Sixth International Conference on Soil Mechanics and Foundation
Engineering, Montreal, Vol. 2, pp. 270–274.
LIU, J. L., YUAN, Z. I., and ZHANG, K. P. (1985). “Cap-Pile-Soil Interaction of Bored Pile Groups,”
Proceedings, Eleventh International Conference on Soil Mechanics and Foundation Engineering, San Francisco, Vol. 3, pp. 1433–1436.
MANSUR, C. I., and HUNTER, A. H. (1970). “Pile Tests—Arkansas River Project,” Journal of the
Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 96,
No. SM6, pp. 1545–1582.
MATLOCK, H., and REESE, L. C. (1960). “Generalized Solution for Laterally Loaded Piles,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers,
Vol. 86, No. SM5, Part I, pp. 63 –91.
MEYERHOF, G. G. (1976). “Bearing Capacity and Settlement of Pile Foundations,” Journal of the
Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 102, No. GT3,
pp. 197–228.
NOTTINGHAM, L. C., and SCHMERTMANN, J. H. (1975). An Investigation of Pile Capacity Design Procedures, Research Report No. D629, Department of Civil Engineering, University of Florida,
Gainesville, FL.
OLSON, R. E., and FLAATE, K. S. (1967). “Pile Driving Formulas for Friction Piles in Sand,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers,
Vol. 93, No. SM6, pp. 279 –296.
O’NEILL, M. W., and REESE, L. C. (1999). Drilled Shafts: Construction Procedure and Design Methods,
FHWA Report No. IF-99-025.
SCHMERTMANN, J. H. (1978). Guidelines for Cone Penetration Test: Performance and Design, Report
FHWA-TS-78-209, Federal Highway Administration, Washington, DC.
SEILER, J. F., and KEENEY, W. D. (1944). “The Efficiency of Piles in Groups,” Wood Preserving News,
Vol. 22, No. 11 (November).
636 Chapter 11: Pile Foundations
SKOV, R., and DENVER, H. (1988). “Time Dependence of Bearing Capacity of Piles,” Proceedings,
Third International Conference on Application of Stress Wave Theory to Piles, Ottawa,
Canada, pp. 879 –889.
SLADEN, J. A. (1992). “The Adhesion Factor: Applications and Limitations,” Canadian Geotechnical
Journal, Vol. 29, No. 2, pp. 323–326.
SVINKIN, M. (1996). Discussion on “Setup and Relaxation in Glacial Sand,” Journal of Geo-technical
Engineering, ASCE, Vol. 22, pp. 319–321.
TERZAGHI, K., PECK, R. B., and MESRI, G. (1996). Soil Mechanics in Engineering Practice, John
Wiley, NY.
VESIC, A. S. (1961). “Bending of Beams Resting on Isotropic Elastic Solids,” Journal of the Engineering
Mechanics Division, American Society of Civil Engineers, Vol. 87, No. EM2, pp. 35–53.
VESIC, A. S. (1969). Experiments with Instrumented Pile Groups in Sand, American Society for Testing and Materials, Special Technical Publication No. 444, pp. 177–222.
VESIC, A. S. (1970). “Tests on Instrumental Piles—Ogeechee River Site,” Journal of the Soil Mechanics
and Foundations Division, American Society of Civil Engineers, Vol. 96, No. SM2, pp. 561–584.
VESIC, A. S. (1977). Design of Pile Foundations, National Cooperative Highway Research Program
Synthesis of Practice No. 42, Transportation Research Board, Washington, DC.
VIJAYVERGIYA, V. N., and FOCHT, J. A., JR. (1972). A New Way to Predict Capacity of Piles in Clay, Offshore Technology Conference Paper 1718, Fourth Offshore Technology Conference, Houston.
WONG, K. S., and TEH, C. I. (1995). “Negative Skin Friction on Piles in Layered Soil Deposit,”
Journal of Geotechnical and Geoenvironmental Engineering, American Society of Civil
Engineers, Vol. 121, No. 6, pp. 457– 465.
WOODWARD, R. J., GARDNER, W. S., and GREER, D. M. (1972). Drilled Pier Foundations, McGraw-Hill,
New York.
12
12.1
Drilled-Shaft Foundations
Introduction
The terms caisson, pier, drilled shaft, and drilled pier are often used interchangeably in
foundation engineering; all refer to a cast-in-place pile generally having a diameter of
about 750 mm or more, with or without steel reinforcement and with or without an enlarged
bottom. Sometimes the diameter can be as small as 305 mm.
To avoid confusion, we use the term drilled shaft for a hole drilled or excavated to
the bottom of a structure’s foundation and then filled with concrete. Depending on the soil
conditions, casings may be used to prevent the soil around the hole from caving in during
construction. The diameter of the shaft is usually large enough for a person to enter for
inspection.
The use of drilled-shaft foundations has several advantages:
1. A single drilled shaft may be used instead of a group of piles and the pile cap.
2. Constructing drilled shafts in deposits of dense sand and gravel is easier than
driving piles.
3. Drilled shafts may be constructed before grading operations are completed.
4. When piles are driven by a hammer, the ground vibration may cause damage to
nearby structures. The use of drilled shafts avoids this problem.
5. Piles driven into clay soils may produce ground heaving and cause previously
driven piles to move laterally. This does not occur during the construction of
drilled shafts.
6. There is no hammer noise during the construction of drilled shafts; there is during
pile driving.
7. Because the base of a drilled shaft can be enlarged, it provides great resistance to
the uplifting load.
8. The surface over which the base of the drilled shaft is constructed can be visually
inspected.
9. The construction of drilled shafts generally utilizes mobile equipment, which, under
proper soil conditions, may prove to be more economical than methods of constructing pile foundations.
10. Drilled shafts have high resistance to lateral loads.
637
638 Chapter 12: Drilled-Shaft Foundations
There are also a couple of drawbacks to the use of drilled-shaft construction. For one
thing, the concreting operation may be delayed by bad weather and always needs close
supervision. For another, as in the case of braced cuts, deep excavations for drilled shafts may
induce substantial ground loss and damage to nearby structures.
12.2
Types of Drilled Shafts
Drilled shafts are classified according to the ways in which they are designed to transfer
the structural load to the substratum. Figure 12.1a shows a drilled straight shaft. It extends
through the upper layer(s) of poor soil, and its tip rests on a strong load-bearing soil layer
or rock. The shaft can be cased with steel shell or pipe when required (as it is with cased,
cast-in-place concrete piles; see Figure 11.4). For such shafts, the resistance to the applied
load may develop from end bearing and also from side friction at the shaft perimeter and
soil interface.
A belled shaft (see Figures 12.1b and c) consists of a straight shaft with a bell at
the bottom, which rests on good bearing soil. The bell can be constructed in the shape
of a dome (see Figure 12.1b), or it can be angled (see Figure 12.1c). For angled bells,
the underreaming tools that are commercially available can make 30 to 45° angles with
the vertical. For the majority of drilled shafts constructed in the United States, the entire
load-carrying capacity is assigned to the end bearing only. However, under certain circumstances, the end-bearing capacity and the side friction are taken into account. In
Europe, both the side frictional resistance and the end-bearing capacity are always taken
into account.
Straight shafts can also be extended into an underlying rock layer. (See
Figure 12.1d.) In the calculation of the load-bearing capacity of such shafts, the end bearing and the shear stress developed along the shaft perimeter and rock interface can be taken
into account.
Soft
soil
Soft
soil
45˚ or
30˚
Rock or
hard soil
(a)
Good
bearing
soil
(b)
Good
bearing
soil
(c)
Rock
0.15 to
0.3 m
(d)
Figure 12.1 Types of drilled shaft: (a) straight shaft; (b) and (c) belled shaft; (d) straight shaft
socketed into rock
12.3 Construction Procedures
12.3
639
Construction Procedures
The most common construction procedure used in the United States involves rotary
drilling. There are three major types of construction methods: the dry method, the casing
method, and the wet method.
Dry Method of Construction
This method is employed in soils and rocks that are above the water table and that will not
cave in when the hole is drilled to its full depth. The sequence of construction, shown in
Figure 12.2, is as follows:
Step 1. The excavation is completed (and belled if desired), using proper drilling tools,
and the spoils from the hole are deposited nearby. (See Figure 12.2a.)
Step 2. Concrete is then poured into the cylindrical hole. (See Figure 12.2b.)
Step 3. If desired, a rebar cage is placed in the upper portion of the shaft. (See
Figure 12.2c.)
Step 4. Concreting is then completed, and the drilled shaft will be as shown in
Figure 12.2d.
Casing Method of Construction
This method is used in soils or rocks in which caving or excessive deformation is likely to
occur when the borehole is excavated. The sequence of construction is shown in Figure 12.3
and may be explained as follows:
Step 1. The excavation procedure is initiated as in the case of the dry method of
construction. (See Figure 12.3a.)
Step 2. When the caving soil is encountered, bentonite slurry is introduced into
the borehole. (See Figure 12.3b.) Drilling is continued until the excavation goes past the caving soil and a layer of impermeable soil or rock is
encountered.
Step 3. A casing is then introduced into the hole. (See Figure 12.3c.)
Step 4. The slurry is bailed out of the casing with a submersible pump. (See
Figure 12.3d.)
Step 5. A smaller drill that can pass through the casing is introduced into the hole,
and excavation continues. (See Figure 12.3e.)
Step 6. If needed, the base of the excavated hole can then be enlarged, using an
underreamer. (See Figure 12.3f.)
Step 7. If reinforcing steel is needed, the rebar cage needs to extend the full length
of the excavation. Concrete is then poured into the excavation and the casing is gradually pulled out. (See Figure 12.3g.)
Step 8. Figure 12.3h shows the completed drilled shaft.
Wet Method of Construction
This method is sometimes referred to as the slurry displacement method. Slurry is used to
keep the borehole open during the entire depth of excavation. (See Figure 12.4.) Following
are the steps involved in the wet method of construction:
640 Chapter 12: Drilled-Shaft Foundations
Surface casing,
if required
Competent,
noncaving
soil
Drop chute
(a)
Competent,
noncaving
soil
(b)
Competent,
noncaving
soil
Competent,
noncaving
soil
(c)
(d)
Figure 12.2 Dry method of construction: (a) initiating drilling; (b) starting concrete pour;
(c) placing rebar cage; (d) completed shaft (After O’Neill and Reese, 1999)
12.3 Construction Procedures
Drilling slurry
Cohesive soil
Cohesive soil
Caving soil
Caving soil
Cohesive soil
Cohesive soil
(a)
(b)
Cohesive soil
Cohesive soil
Caving soil
Caving soil
Cohesive soil
Cohesive soil
(c)
(d)
Figure 12.3 Casing method of construction: (a) initiating drilling; (b) drilling with
slurry; (c) introducing casing; (d) casing is sealed and slurry is being removed from
interior of casing; (e) drilling below casing; (f) underreaming; (g) removing casing;
(h) completed shaft (After O’Neill and Reese, 1999)
641
642 Chapter 12: Drilled-Shaft Foundations
Competent soil
Competent soil
Caving soil
Caving soil
Competent soil
Competent soil
(e)
(f)
Level of
fluid concrete
Drifting fluid
forced from
space between
casing and soil
Competent soil
Competent soil
Caving soil
Caving soil
Competent soil
Competent soil
(g)
Figure 12.3 (Continued)
(h)
12.3 Construction Procedures
Drilling slurry
Cohesive soil
Caving soil
643
Cohesive soil
Caving soil
(b)
(a)
Cohesive soil
Cohesive soil
Sump
Caving soil
Caving soil
(c)
(d)
Figure 12.4 Slurry method of construction: (a) drilling to full depth with slurry; (b) placing rebar
cage; (c) placing concrete; (d) completed shaft (After O’Neill and Reese, 1999)
Step 1. Excavation continues to full depth with slurry. (See Figure 12.4a.)
Step 2. If reinforcement is required, the rebar cage is placed in the slurry. (See
Figure 12.4b.)
Step 3. Concrete that will displace the volume of slurry is then placed in the drill
hole. (See Figure 12.4c.)
Step 4. Figure 12.4d shows the completed drilled shaft.
Figure 12.5 shows a drilled shaft under construction using the dry method. The construction of a drilled shaft using the wet method is shown in Figure 12.6. A typical auger,
a reinforcement cage, and a typical clean-out bucket are shown in Figure 12.7.
644 Chapter 12: Drilled-Shaft Foundations
Figure 12.5 Drilled shaft
construction using the dry
method (Courtesy of Sanjeev
Kumar, Southern Illinois
University, Carbondale,
Illinois)
Figure 12.6 Drilled shaft construction using wet method (Courtesy of Khaled Sobhan,
Florida Atlantic Univetsity, Boca Raton, Florida)
12.4 Other Design Considerations
(a)
645
(b)
(c)
Figure 12.7 Drilled shaft construction: (a) A typical auger; (b) a reinforcement cage; (c) a cleanout bucket (Courtesy of Khaled Sobhan, Florida Atlantic Univetsity, Boca Raton, Florida)
12.4
Other Design Considerations
For the design of ordinary drilled shafts without casings, a minimum amount of vertical
steel reinforcement is always desirable. Minimum reinforcement is 1% of the gross
cross-sectional area of the shaft. For drilled shafts with nominal reinforcement, most
building codes suggest using a design concrete strength, fc , on the order of fcr >4. Thus,
the minimum shaft diameter becomes
fc 5 0.25fcr 5
Qw
Qw
5
p 2
A gs
Ds
4
646 Chapter 12: Drilled-Shaft Foundations
or
Ds 5
Qw
p
¢ ≤ (0.25)fcr
ç 4
Qu
Å fcr
5 2.257
(12.1)
where
Ds 5 diameter of the shaft
fcr 5 28-day concrete strength
Qw 5 working load of the drilled shaft
A gs 5 gross cross-sectional area of the shaft
If drilled shafts are likely to be subjected to tensile loads, reinforcement should be continued for the entire length of the shaft.
Concrete Mix Design
The concrete mix design for drilled shafts is not much different from that for any other
concrete structure. When a reinforcing cage is used, consideration should be given to the
ability of the concrete to flow through the reinforcement. In most cases, a concrete slump
of about 15.0 mm (6 in.) is considered satisfactory. Also, the maximum size of the aggregate should be limited to about 20 mm (0.75 in.)
12.5
Load Transfer Mechanism
The load transfer mechanism from drilled shafts to soil is similar to that of piles, as
described in Section 11.5. Figure 12.8 shows the load test results of a drilled shaft, conducted in a clay soil by Reese et al. (1976). The shaft (Figure 12.8a) had a diameter of
762 mm and a depth of penetration of 6.94 m. Figure 12.8b shows the load-settlement
curves. It can be seen that the total load carried by the drilled shaft was 1246 kN. The
load carried by side resistance was about 800 kN, and the rest was carried by point bearing. It is interesting to note that, with a downward movement of about 6 mm, full side
resistance was mobilized. However, about 25 mm of downward movement was required
for mobilization of full point resistance. This situation is similar to that observed in the
case of piles. Figure 12.8c shows the average load-distribution curves for different stages
of the loading.
12.6
Estimation of Load-Bearing Capacity
The ultimate load-bearing capacity of a drilled shaft (see Figure 12.9) is
Qu 5 Qp 1 Qs
(12.2)
12.6 Estimation of Load-Bearing Capacity
647
Steel loading
plate
0.762 m
1200
Load (kN)
762
mm
6.94 m
Total
800
Sides
400
0
114 mm
Base
0
Bottom
hole load cell
(a)
0
0
6
12 18 24
Mean settlement (mm)
30
(b)
400
Load (kN)
800
1200 1400
Depth (m)
1.5
3.0
4.5
6.0
7.5
(c)
Figure 12.8 Load test results of
Reese et al. (1976) on a drilled shaft:
(a) dimensions of the shaft; (b) plot
of base, sides, and total load with
mean settlement; (c) plot of loaddistribution curve with depth
where
Qu 5 ultimate load
Qp 5 ultimate load-carrying capacity at the base
Qs 5 frictional (skin) resistance
The ultimate base load Qp can be expressed in a manner similar to the way it is
expressed in the case of shallow foundations [Eq. (3.19)], or
1
Qp 5 A p acrNcFcsFcdFcc 1 qrNqFqsFqdFqc 1 grNgFgsFgdFgc b
(12.3)
2
648 Chapter 12: Drilled-Shaft Foundations
Qu
Qu
Qs
Qs
z
z
L1
DS
Db
Qp
(a)
L
Soil
␾
c
L L1
Db Ds
Qp
Soil
␾
c
(b)
Figure 12.9 Ultimate bearing capacity of drilled shafts: (a) with bell and (b) straight shaft
where
cr 5 cohesion
Nc , Nq , Ng 5 bearing capacity factors
Fcs , Fqs , Fgs 5 shape factors
Fcd , Fqd , Fgd 5 depth factors
Fcc , Fqc , Fgc 5 compressibility factors
gr 5 effective unit weight of soil at the base of the shaft
qr 5 effective vertical stress at the base of the shaft
p
A p 5 area of the base 5 D2b
4
In most instances, the last term (the one containing Ng) is neglected, except in the
case of a relatively short drilled shaft. With this assumption, we can write
Qu 5 A p (c rNcFcsFcdFcc 1 qNqFqsFqdFqc ) 1 Qs
(12.4)
The procedure to estimate the ultimate capacity of drilled shafts in granular and cohesive
soil is described in the following sections.
12.7
Drilled Shafts in Granular Soil: Load-Bearing Capacity
Estimation of Qp
For a drilled shaft with its base located on a granular soil (that is, c r 5 0), the net ultimate
load-carrying capacity at the base can be obtained from Eq. (12.4) as
Qp(net) 5 A p 3q r (Nq 2 1)FqsFqdFqc4
(12.5)
649
12.7 Drilled Shafts in Granular Soil: Load-Bearing Capacity
The bearing capacity factor, Nq, for various soil friction angles (fr ) can be taken from
Table 3.3. It is also given in Table 12.1. Also,
Fqs 5 1 1 tan fr
(12.6)
Fqd 5 1 1 C tan21 a
L
b
Db
(')'*
radian
(12.7)
C 5 2 tan fr (1 2 sin fr ) 2
(12.8)
The variations of Fqs and C with fr are given in Table 12.1.
According to Chen and Kulhawy (1994), Fqc can be calculated in the following
manner.
Step 1. Calculate the critical rigidity index as
Icr 5 0.5 exp c2.85 cota45 2
fr
bd
2
(12.9)
where Icr 5 critical rigidity index (see Table 12.1).
Table 12.1 Variation of Nq, Fqs, C, Icr, ms, and n with fr
Soil friction
angle, f9
(deg)
Nq
(Table 3.3)
Fqs
[Eq. (12.6)]
C
[Eq. (12.8)]
Icr
[Eq. (12.9)]
ms
[Eq. (12.13)]
n
[Eq. (12.15)]
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
10.66
11.85
13.20
14.72
16.44
18.40
20.63
23.18
26.09
29.44
33.30
37.75
42.92
48.93
55.96
64.20
73.90
85.38
99.02
115.31
134.88
1.466
1.488
1.510
1.532
1.554
1.577
1.601
1.625
1.649
1.675
1.700
1.727
1.754
1.781
1.810
1.839
1.869
1.900
1.933
1.966
2.000
0.311
0.308
0.304
0.299
0.294
0.289
0.283
0.276
0.269
0.262
0.255
0.247
0.239
0.231
0.223
0.214
0.206
0.197
0.189
0.180
0.172
43.84
47.84
52.33
57.40
63.13
69.63
77.03
85.49
95.19
106.37
119.30
134.33
151.88
172.47
196.76
225.59
259.98
301.29
351.22
412.00
486.56
0.100
0.115
0.130
0.145
0.160
0.175
0.190
0.205
0.220
0.235
0.250
0.265
0.280
0.295
0.310
0.325
0.340
0.355
0.370
0.385
0.400
0.00500
0.00475
0.00450
0.00425
0.00400
0.00375
0.00350
0.00325
0.00300
0.00275
0.00250
0.00225
0.00200
0.00175
0.00150
0.00125
0.00100
0.00075
0.00050
0.00025
0.00000
650 Chapter 12: Drilled-Shaft Foundations
Step 2. Calculate the reduced rigidity index as
Irr 5
Ir
1 1 Ir D
(12.10)
where
Ir 5 soil rididity index 5
Es
2(1 1 ms )q r tan fr
(12.11)
in which
Es 5 drained modulus of elasticity of soil 5 mpa
pa 5 atmospheric pressure( < 100 kN>m2 )
(12.12)
100 to 200 (loose soil)
m 5 c200 to 500 (medium dense soil)
500 to 1000(dense soil)
ms 5 Poisson’s ratio of soil 5 0.1 1 0.3a
fr 2 25
b
20
(for 25° # f r # 45°) (see Table12.1)
D5n
n 5 0.005a1 2
qr
pa
(12.13)
(12.14)
fr 2 25
b (see Table 12.1)
20
(12.15)
Step 3. If Irr $ Icr, then
Fqc 5 1
(12.16)
However, if Irr , Icr, then
Fqc 5 exp e (23.8 tan fr ) 1 c
(3.07 sin fr ) ( log 10 2Irr )
df
1 1 sin fr
(12.17)
The magnitude of Qp(net) also can be reasonably estimated from a
relationship based on the analysis of Berezantzev et al. (1961) that can be
expressed as
Qp(net) 5 A pq r (vN *q 2 1)
(12.18)
where
N *q 5 bearing capacity factor 5 0.21e0.17fr (See Table 12.2)
v 5 correction factor 5 f(L>Db )
(12.19)
In Eq. (12.19), f r is in degrees. The variation of v with L>Db is given
in Figure 12.10.
12.7 Drilled Shafts in Granular Soil: Load-Bearing Capacity
651
Table 12.2 Variation of N q* with fr [Eq. (12.19)]
fr (deg)
N*q
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
14.72
17.45
20.68
24.52
29.06
34.44
40.83
48.39
57.36
67.99
80.59
95.52
113.22
134.20
159.07
188.55
223.49
264.90
313.99
372.17
441.14
1.0
0.9
L/Db 0.8
5
␻ 0.7
10
15
0.6
20
25
0.5
0.4
26
28
30
32
34
36
Soil friction angle, ␾ (deg)
38
40
Figure 12.10 Variation of v
with fr and L>Db
652 Chapter 12: Drilled-Shaft Foundations
Estimation of Qs
The frictional resistance at ultimate load, Qs , developed in a drilled shaft may be calculated as
L1
Qs 5 3 pfdz
(12.20)
0
where
p 5 shaft perimeter 5 pDs
f 5 unit frictional (or skin) resistance 5 Ksor tan dr
(12.21)
K 5 earth pressure coefficient < Ko 5 1 2 sin fr
(12.22)
sor 5 effective vertical stress at any depth z
Thus,
L1
L1
Qs 5 3 pfdz 5 pDs (1 2 sin fr ) 3 sor tan dr dz
0
(12.23)
0
The value of sor will increase to a depth of about 15Ds and will remain constant thereafter,
as shown in Figure 11.16.
For cast-in-pile concrete and good construction techniques, a rough interface
develops and, hence, dr>fr may be taken to be one. With poor slurry construction,
dr>fr < 0.7 to 0.8.
Allowable Net Load, Qall (net)
An appropriate factor of safety should be applied to the ultimate load to obtain the net
allowable load, or
Qall(net) 5
12.8
Qp(net) 1 Qs
FS
(12.24)
Load-Bearing Capacity Based on Settlement
On the basis of a database of 41 loading tests, Reese and O’Neill (1989) proposed a
method for calculating the load-bearing capacity of drilled shafts that is based on settlement. The method is applicable to the following ranges:
1.
2.
3.
4.
Shaft diameter: Ds 5 0.52 to 1.2 m
Bell depth: L 5 4.7 to 30.5 m
Field standard penetration resistance: N60 5 5 to 60
Concrete slump 5 100 to 225 mm
12.8 Load-Bearing Capacity Based on Settlement
653
Qu
1.5 m (5 ft) noncontributing
zone (cohesive soil only)
i1
i2
Li
Ds
L
fi p Li
ii
L1
Noncontributing zones:
Length Ds (cohesive soil only)
iN
Db
No side-load transfer
permitted on perimeter
of bell
qp Ap
Figure 12.11 Development of Eq. (12.25)
Reese and O’Neill’s procedure (see Figure 12.11) gives
N
Qu(net) 5 a fipDLi 1 qpA p
(12.25)
i51
where
fi 5 ultimate unit shearing resistance in layer i
p 5 perimeter of the shaft 5 pDs
qp 5 unit point resistance
A p 5 area of the base 5 (p>4)D2b
fi 5 b1sozi
r , b2
(12.26)
r 5 vertical effective stress at the middle of layer i.
where sozi
b1 5 b3 2 b4z0.5
i (for 0.25 # b1 # 1.2)
(12.27)
654 Chapter 12: Drilled-Shaft Foundations
The units for fi, zi, and sozi
r and the magnitude of b2, b3, and b4 in the SI system are
Item
SI
fi
zi
sozi
r
b2
b3
b4
kN>m2
m
kN>m2
192 kN>m2
1.5
0.244
The point bearing capacity is
qp 5 b5N60 # b6
3for Db , 1.27 m4
(12.28)
where N60 5 field standard penetration number within a distance of 2Db below the base of
the drilled shaft.
The magnitudes of b5 and b6 and the unit of qp in the SI system are given here.
Item
SI
b5
b6
qp
57.5
4310 kN>m2
kN>m2
If Db is equal to or greater than 1.27 m, excessive settlement may occur. In that case, qp
may be replaced by qpr or
qpr 5
1.27
q
Db (m) p
(12.29)
Based on the desired level of settlement, Figures 12.12 and 12.13 may now be used to calculate the allowable load, Qall(net). Note that the trend lines given in these figures is the average
of all test results.
More recently, Rollins et al. (2005) have modified Eq. (12.27) for gravelly sands as
follows:
For sand with 25 to 50% gravel,
(for 0.25 # b1 # 1.8)
b1 5 b7 2 b8z0.75
i
(12.30)
12.8 Load-Bearing Capacity Based on Settlement
655
2.0
End bearing
Ultimate end bearing, qp Ap
1.6
1.2
Trend
line
0.8
0.4
0
0
2
4
6
8
Settlement of base
(%)
Diameter of base, Db
10
12
Figure 12.12
Normalized base-load
transfer versus settlement
in sand
Side-load transfer
Ultimate side-load transfer, fi p Li
1.2
1.0
Trend
line
0.8
0.6
0.4
0.2
0
0
0.4
0.8
1.2
1.6
Settlement
(%)
Diameter of shaft, Ds
2.0
Figure 12.13 Normalized side-load
transfer versus settlement in sand
For sand with more than 50% gravel,
b1 5 b9e2b10zi (for 0.25 # b1 # 3.0)
(12.31)
The magnitudes of b7, b8, b9, and b10 and the unit of zi in the SI system are given here.
656 Chapter 12: Drilled-Shaft Foundations
Item
SI
b7
b8
b9
b10
zi
2.0
0.15
3.4
2 0.085
m
Figure 12.14 provides the normalized side-load transfer trend based on the desired
level of settlement for gravelly sand and gravel.
Side-load transfer
Ultimate side-load transfer, fi p Li
1.0
0.8
0.6
Trend
line
0.4
0.2
0
0
0.4
0.8
1.2
1.6
Settlement of base
(%)
Diameter of shaft, Ds
(a)
2.0
2.4
2.0
2.4
Side-load transfer
Ultimate side-load transfer, fi p Li
1.0
0.8
Trend
line
0.6
0.4
0.2
0
0
0.4
0.8
1.2
1.6
Settlement of base
(%)
Diameter of shaft, Ds
(b)
Figure 12.14 Normalized side-load transfer versus settlement: (a) gravelly sand
(gravel 25–50%) and (b) gravel (more than 50%)
12.8 Load-Bearing Capacity Based on Settlement
657
Example 12.1
A soil profile is shown in Figure 12.15. A point bearing drilled shaft with a bell is
placed in a layer of dense sand and gravel. Determine the allowable load the drilled
shaft could carry. Use Eq. (12.5) and a factor of safety of 4. Take Ds 5 1 m and
Db 5 1.75 m. For the dense sand layer, fr 5 36°; Es 5 500pa . Ignore the frictional resistance of the shaft.
Solution
We have
Qp(net) 5 A p 3qr(Nq 2 1)FqsFqdFqc4
and
qr 5 (6) (16.2) 1 (2) (19.2) 5 135.6 kN>m2
For fr 5 36°, from Table 12.1, Nq 5 37.75. Also,
Fqs 5 1.727
and
Fqd 5 1 1 C tan21 a
L
≤
Db
5 1 1 0.247 tan21 ¢
8
≤ 5 1.335
1.75
Qu
6m
Ds
Loose sand
␥ 16.2 kN/m3
Dense sand and
gravel
␥ 19.2 kN/m3
␾ 36˚
2m
Db
Figure 12.15 Allowable load of drilled shaft
658 Chapter 12: Drilled-Shaft Foundations
From Eq. (12.9),
Icr 5 0.5 expB2.85 cot¢ 45 2
fr
≤ R 5 134.3 (See Table 12.1)
2
From Eq. (12.12), Es 5 mpa . With m 5 500, we have
Es 5 (500) (100) 5 50,000 kN>m2
From Eq. (12.13) and Table 12.1,
ms 5 0.265
So
Ir 5
Es
50,000
5
5 200.6
2(1 1 ms ) (qr) (tan fr)
2(1 1 0.265) (135.6) (tan 36)
From Eq. (12.10),
Irr 5
Ir
1 1 Ir D
with
D5n
qr
135.6
5 0.00225¢
≤ 5 0.0031
pa
100
it follows that
Irr 5
200.6
5 123.7
1 1 (200.6) (0.0031)
Irr is less than Icr . So, from Eq. (12.17),
Fqc 5 exp b (23.8 tan fr) 1 B
(3.07 sin fr) (log 10 2Irr )
Rr
1 1 sin fr
5 exp b (23.8 tan 36) 1 B
(3.07 sin 36) log(2 3 123.7)
R r 5 0.958
1 1 sin 36
Hence,
p
Qp(net) 5 B¢ ≤ (1.75) 2R (135.6) (37.75 2 1) (1.727) (1.335) (0.958) 5 26,474 kN
4
and
Q p(all) 5
Qp(net)
FS
5
26,474
< 6619 kN
4
■
12.8 Load-Bearing Capacity Based on Settlement
659
Example 12.2
Solve Example 12.1 using Eq. (12.18).
Solution
Equation (12.18) asserts that
Qp(net) 5 A pqr(vN q* 2 1)
We have (also see Table 12.2)
N q* 5 0.21e0.17fr 5 0.21e (0.17)(36) 5 95.52
and
L
8
5
5 4.57
Db
1.75
From Figure 12.10, for fr 5 36° and L>Db 5 4.57, the value of v is about 0.83. So
p
Qp(net) 5 B ¢ ≤ (1.75) 2 R (135.6) 3(0.83) (95.52) 2 14 5 25,532 kN
4
and
Qp(all) 5
25,532
5 6383 kN
4
■
Example 12.3
A drilled shaft is shown in Figure 12.16. The uncorrected average standard penetration number (N60 ) within a distance of 2Db below the base of the shaft is about 30.
Determine
a. The ultimate load-carrying capacity
b. The load-carrying capacity for a settlement of 12mm. Use Eq. (12.30).
Solution
Part a
From Eqs. (12.26) and (12.27),
fi 5 b1sozi
r
and
b1 5 2.0 2 0.15z0.75
For this problem, zi 5 6>2 5 3 m, so
b 5 2 2 (0.15) (3) 0.75 5 1.658
660 Chapter 12: Drilled-Shaft Foundations
Loose sandy gravel
␥ 16 kN/m3
6m
1m
1m
1.5 m
Dense sandy gravel
␥ 19 kN/m3
N60 30
Figure 12.16 Drilled shaft supported by a dense layer of sandy gravel
and
sozi
r 5 gzi 5 (16) (3) 5 48 kN>m2
Thus,
fi 5 (48) (1.658) 5 79.58 kN>m2
and
Sfi p DLi 5 (79.58) (p 3 1) (6) 5 1500 kN
From Eq. (12.28),
qp 5 57.5N60 5 (57.5) (30) 5 1725 kN>m2
Note that Db is greater than 1.27. So we will use Eq. (12.29a).
qpr 5 a
1.27
1.27
bqp 5 ¢
≤ (1725) < 1461 kN>m2
Db
1.5
Now
qprA p 5 (14.61) ¢
p
3 1.52 ≤ < 2582 kN
4
Hence,
Qult(net) 5 qprA p 1 Sfi p DLi 5 2582 1 1500 5 4082 kN
12.9 Drilled Shafts in Clay: Load-Bearing Capacity
661
Part b
We have
Allowable settlement
12
5
5 0.12 5 1.2%
Ds
(1.0) (1000)
The trend line in Figure 12.14a shows that, for a normalized settlement of 1.2%, the
normalized load is about 0.8 . Thus, the side-load transfer is (0.8) (1500) < 1200 kN.
Similarly,
Allowable settlement
12
5
5 0.008 5 0.8%
Db
(1.5) (1000)
The trend line shown in Figure 12.12 indicates that, for a normalized settlement of
1.4%, the normalized base load is 0.317. So the base load is (0.317) (2582) 5
818.5 kN. Hence, the total load is
Q 5 1200 1 818.5 < 2018.5 kN
12.9
■
Drilled Shafts in Clay: Load-Bearing Capacity
For saturated clays with f 5 0, the bearing capacity factor Nq in Eq. (12.4) is equal to
unity. Thus, for this case,
Qp(net) < A pcuNcFcsFcdFcc
(12.32)
where cu 5 undrained cohesion.
Assuming that L > 3Db , we can rewrite Eq. (12.32) as
Qp(net) 5 A pcuN *c
(12.33)
where N c* 5 NcFcsFcdFcc 5 1.333(ln Ir ) 1 14 in which Ir 5 soil rigidity index.
(12.34)
The soil rigidity index was defined in Eq. (12.11). For f 5 0,
Ir 5
Es
3cu
(12.35)
O’Neill and Reese (1999) provided an approximate relationship between cu and
Es>3cu . This relationship is shown in Figure 12.17. For all practical purposes, if cu>pa is
662 Chapter 12: Drilled-Shaft Foundations
300
250
200
Based on O'Neill
and Reese (1999)
Es
150
3cu
100
50
0
0
0.5
1.0
cu
pa
1.5
2.0
Figure 12.17 Approximate
Es
variation of
with cu>pa
3cu
(Note: pa 5 atmospheric pressure) :
(Based on O’Neill and Reese, 1999)
equal to or greater than unity (pa 5 atmospheric pressure < 100 kN>m2), then the magnitude of N c* can be taken to be 9.
Experiments by Whitaker and Cooke (1966) showed that, for belled shafts, the full
value of N c* 5 9 is realized with a base movement of about 10 to 15% of Db . Similarly,
for straight shafts (Db 5 Ds ), the full value of N c* 5 9 is obtained with a base movement
of about 20% of Db .
The expression for the skin resistance of drilled shafts in clay is similar to
Eq. (11.55), or
L5L1
Qs 5 a a*cup DL
(12.36)
L50
Kulhawy and Jackson (1989) reported the field-test result of 106 straight drilled
shafts—65 in uplift and 41 in compression. The best correlation obtained from the
results is
a* 5 0.21 1 0.25¢
pa
≤ <1
cu
(12.37)
where pa 5 atmospheric pressure < 100 kN>m2.
So, conservatively, we may assume that
a* 5 0.4
(12.38)
12.10 Load-Bearing Capacity Based on Settlement
12.10
663
Load-Bearing Capacity Based on Settlement
Reese and O’Neill (1989) suggested a procedure for estimating the ultimate and allowable
(based on settlement) bearing capacities for drilled shafts in clay. According to this procedure, we can use Eq. (12.25) for the net ultimate load, or
n
Qult(net) 5 a fip DLi 1 qpA p
i51
The unit skin friction resistance can be given as
fi 5 ai*cu(i)
(12.39)
The following values are recommended for a*i :
a*i 5 0 for the top 1.5 m (5 ft) and bottom 1 diameter, Ds, of the drilled shaft. (Note: If
Db . Ds , then a* 5 0 for 1 diameter above the top of the bell and for the peripheral area of the bell itself.)
a*i 5 0.55 elsewhere.
The expression for qp (point load per unit area) can be given as
qp 5 6cub a1 1 0.2
L
b # 9cub # 40pa
Db
(12.40)
where
cub 5 average undrained cohesion within a vertical distance of 2Db below the base
pa 5 atmospheric pressure
If Db is large, excessive settlement will occur at the ultimate load per unit area, qp,
as given by Eq. (12.40). Thus, for Db . 1.91 m, qp may be replaced by
qpr 5 Frqp
(12.41)
2.5
#1
c1Db 1 c2
(12.42)
where
Fr 5
The relations for c1 and c2 along with the unit of Db in the SI system are given in
Table 12.3.
Figures 12.18 and 12.19 may now be used to evaluate the allowable load-bearing
capacity, based on settlement. (Note that the ultimate bearing capacity in Figure 12.18 is
qp , not qpr .) To do so,
664 Chapter 12: Drilled-Shaft Foundations
Step 1. Select a value of settlement, s.
N
Step 2. Calculate a fip DLi and qpA p .
i51
Step 3. Using Figures 12.18 and 12.19 and the calculated values in Step 2, determine the side load and the end bearing load.
Step 4. The sum of the side load and the end bearing load gives the total allowable load.
Table 12.3 Relationships for c1 and c2
Item
SI
c1 5 2.78 3 1024 1 8.26 3 1025 a
c1
c2
c2 5 0.0653cub (kN>m2 )4 0.5
(0.5 # c2 # 1.5)
Db
mm
L
b # 5.9 3 1024
Db
1.2
Side-load transfer
Ultimate side-load transfer, fi p Li
1.0
0.8
Trend
line
0.6
0.4
0.2
0
0
0.4
0.8
1.2
1.6
Settlement
(%)
Diameter of shaft, Ds
2.0
Figure 12.18 Normalized side-load transfer versus settlement in cohesive soil
12.10 Load-Bearing Capacity Based on Settlement
665
1.0
End bearing
Ultimate end bearing, qp Ap
0.8
0.6
Trend
line
0.4
0.2
0
0
2
4
6
Settlement of base
(%)
Diameter of base, Db
8
10
Figure 12.19 Normalized base-load transfer versus settlement in cohesive soil
Example 12.4
Figure 12.20 shows a drilled shaft without a bell. Here, L1 5 8.23 m, L2 5 2.59 m,
Ds 5 1.0 m, cu(1) 5 50 kN>m2, and cu(2) 5 108.75 kN>m2. Determine
a. The net ultimate point bearing capacity
b. The ultimate skin resistance
c. The working load, Qw (FS 5 3)
Use Eqs. (12.33), (12.36), and (12.38).
Solution
Part a
From Eq. (12.33),
p
Qp(net) 5 A pcuN c* 5 A pcu(2)N c* 5 B ¢ ≤ (1) 2 R (108.75) (9) 5 768.7 kN
4
(Note: Since cu(2)>pa . 1, N c* < 9.)
666 Chapter 12: Drilled-Shaft Foundations
Clay
L1
cu (1)
Ds
Clay
L2
cu (2)
Figure 12.20 A drill shaft without a bell
Part b
From Eq. (12.36),
Qs 5 Sa*cupDL
From Eq. (12.38),
a* 5 0.4
p 5 pDs 5 (3.14) (1.0) 5 3.14 m
and
Qs 5 (0.4) (3.14) 3(50 3 8.23) 1 (108.75 3 2.59) 4 5 871 kN
Part c
Qw 5
Qp(net) 1 Qs
FS
5
768.7 1 871
5 546.6 kN
3
■
Example 12.5
A drilled shaft in a cohesive soil is shown in Figure 12.21. Use Reese and O’Neill’s
method to determine the following.
a. The ultimate load-carrying capacity.
b. The load-carrying capacity for an allowable settlement of 12 mm.
12.10 Load-Bearing Capacity Based on Settlement
Clay
cu(1) 40 kN/m2
0.76 m
3m
6m
3m
Clay
cu(2) 60 kN/m2
1.5 m
Clay
cu 145 kN/m2
1.2 m
Figure 12.21 A drilled shaft in layered clay
Solution
Part a
From Eq. (12.39),
fi 5 ai*cu(i)
From Figure 12.21,
DL1 5 3 2 1.5 5 1.5 m
DL2 5 (6 2 3) 2 Ds 5 (6 2 3) 2 0.76 5 2.24 m
cu(1) 5 40 kN>m2
and
cu(2) 5 60 kN>m2
Hence,
SfipDLi 5 Sai*cu(i)pDLi
5 (0.55) (40) (p 3 0.76) (1.5) 1 (0.55) (60) (p 3 0.76) (2.24)
5 255.28 kN
667
668 Chapter 12: Drilled-Shaft Foundations
Again, from Eq. (12.40),
qp 5 6cub ¢1 1 0.2
L
6 1 1.5
≤ 5 (6) (145) B1 1 0.2¢
≤ R 5 1957.5 kN>m2
Db
1.2
A check reveals that
qp 5 9cub 5 (9) (145) 5 1305 kN>m2 , 1957.5 kN>m2
So we use qp 5 1305 kN>m2
p
p
qpA p 5 qp ¢ D2b ≤ 5 (1305) B ¢ ≤ (1.2) 2 R 5 1475.9 kN
4
4
Hence,
Qult 5 Sai*cu(i)pDLi 1 qpA p 5 255.28 1 1475.9 < 1731 kN
Part b
We have
12
Allowable settlement
5
5 0.0158 5 1.58%
Ds
(0.76) (1000)
The trend line shown in Figure 12.18 indicates that, for a normalized settlement of
1.58%, the normalized side load is about 0.9. Thus, the side load is
(0.9) (SfipDLi ) 5 (0.9) (255.28) 5 229.8 kN
Again,
Allowable settlement
12
5
5 0.01 5 1.0%
Db
(1.2) (1000)
The trend line shown in Figure 12.19 indicates that, for a normalized settlement of
1.0%, the normalized end bearing is about 0.63, so
Base load 5 (0.63) (qpA p ) 5 (0.63) (1475.9) 5 929.8 kN
Thus, the total load is
Q 5 229.8 1 929.8 5 1159.6 kN
12.11
■
Settlement of Drilled Shafts at Working Load
The settlement of drilled shafts at working load is calculated in a manner similar to that
outlined in Section 11.15. In many cases, the load carried by shaft resistance is small
compared with the load carried at the base. In such cases, the contribution of s3 may be
ignored. (Note that in Eqs. (11.74) and (11.75) the term D should be replaced by Db for
drilled shafts.)
12.11 Settlement of Drilled Shafts at Working Load
669
Example 12.6
Refer to Figure 12.20. Given: L1 5 8 m, L2 5 3 m, Ds 5 1.5 m, cu(1) 5 50 kN>m2,
cu(2) 5 105 kN>m2, and working load Qw 5 1005 kN. Estimate the elastic settlement
at the working load. Use Eqs. (11.73), (11.75), and (11.76). Take j 5 0.65,
Ep 5 21 3 106 kN>m2, Es 5 14,000 kN>m2, ms 5 0.3, and Qwp 5 250 kN.
Solution
From Eq. (11.73),
se(1) 5
(Qwp 1 jQ ws )L
A pEp
Now,
Qws 5 1005 2 250 5 755 kN
so
se(1) 5
3250 1 (0.65) (755)4 (11)
p
¢ 3 1.52 ≤ (21 3 106 )
4
5 0.00022 m 5 0.22 mm
From Eq. (11.75),
se(2) 5
QwpCp
Dbqp
From Table 11.13, for stiff clay, Cp < 0.04; also,
qp 5 cu(b)N c* 5 (105) (9) 5 945 kN>m2
Hence,
se(2) 5
(250) (0.04)
5 0.0071 m 5 7.1 mm
(1.5) (945)
Again, from Eqs. (11.76) and (11.77),
se(3) 5 ¢
Qws Ds
≤ ¢ ≤ (1 2 m2s )Iws
pL
Es
where
L
11
5 2 1 0.35
5 2.95
Å Ds
Å 1.5
Iws 5 2 1 0.35
se(3) 5 B
755
1.5
R¢
≤ (1 2 0.32 ) (2.95) 5 0.0042 m 5 4.2 mm
(p 3 1.5) (11)
14,000
670 Chapter 12: Drilled-Shaft Foundations
The total settlement is
se 5 se(1) 1 se(2) 1 se(3) 5 0.22 1 7.1 1 4.2 < 11.52 mm
12.12
■
Lateral Load-Carrying Capacity—Characteristic
Load and Moment Method
Several methods for analyzing the lateral load-carrying capacity of piles, as well as the
load-carrying capacity of drilled shafts, were presented in Section 11.16; therefore, they
will not be repeated here. In 1994, Duncan et al. developed a characteristic load method
for estimating the lateral load capacity for drilled shafts that is fairly simple to use. We
describe this method next.
According to the characteristic load method, the characteristic load Qc and moment
Mc form the basis for the dimensionless relationship that can be given by the following
correlations:
Characteristic Load
Qc 5 7.34D2s (EpRI ) ¢
Qc 5 1.57D2s (EpRI ) ¢
cu 0.68
≤
EpRI
grDsfrKp
EpRI
(for clay)
(12.43)
(for sand)
(12.44)
(for clay)
(12.45)
(for sand)
(12.46)
0.57
≤
Characteristic Moment
Mc 5 3.86D3s (EpRI ) ¢
Mc 5 1.33D3s (EpRI ) ¢
cu 0.46
≤
EpRI
grDsfrKp
EpRI
0.40
≤
In these equations,
Ds 5 diameter of drilled shafts
Ep 5 modulus of elasticity of drilled shafts
RI 5 ratio of moment of inertia of drilled shaft section to moment of inertia of a solid
section (Note: RI 5 1 for uncracked shaft without central void)
gr 5 effective unit weight of sand
fr 5 effective soil friction angle (degrees)
Kp 5 Rankine passive pressure coefficient 5 tan2 (45 1 fr>2)
Deflection Due to Load Qg Applied at the Ground Line
Figures 12.22 and 12.23 give the plot of Qg>Qc versus xo>Ds for drilled shafts in sand
and clay due to the load Qg applied at the ground surface. Note that xo is the ground
12.12 Lateral Load-Carrying Capacity—Characteristic Load and Moment Method
0.050
0.050
0.045
0.045
671
Qg
Fixed head
Qc
0.030
0.030
Qg
Free head
Qc
Mg
Mc
Qg
Qc
Mg
Mc
0.015
xo
Ds
0.015
Qg
Qg
Free
Fixed
Qc
Qc
0.005
0.010
0.020
0.0065
0.0091
0.0135
0.0133
0.0197
0.0289
Mg
Mc
0.0024
0.0048
0.0074
0
0
0
0.05
0.10
0.15
xo
Ds
Figure 12.22 Plot of
Qg
Qc
and
Mg
Mc
versus
xo
in clay
Ds
line deflection. If the magnitudes of Qg and Qc are known, the ratio Qg>Qc can be calculated. The figure can then be used to estimate the corresponding value of xo>Ds and,
hence, xo .
Deflection Due to Moment Applied at the Ground Line
Figures 12.22 and 12.23 give the variation plot of Mg>Mc with xo>Ds for drilled shafts in sand
and clay due to an applied moment Mg at the ground line. Again, xo is the ground line deflection. If the magnitudes of Mg , Mc , and Ds are known, the value of xo can be calculated with
the use of the figure.
672 Chapter 12: Drilled-Shaft Foundations
0.015
0.015
Mg
Mc
Qg
Fixed head
Qc
0.010
0.010
Qg
Free head
Qc
Qg
Qc
0.005
Mg
Mc
xo
Ds
Qg
Free
Qc
Qg
Fixed
Qc
Mg
Mc
0.005
0.010
0.020
0.0013
0.0021
0.0033
0.0028
0.0049
0.0079
0.0009
0.0019
0.0032
0.005
0
0
0
0.05
0.10
0.15
xo
Ds
Figure 12.23 Plot of
Qg
Qc
and
Mg
Mc
versus
xo
in sand
Ds
Deflection Due to Load Applied Above the Ground Line
When a load Q is applied above the ground line, it induces both a load Qg 5 Q and a
moment Mg 5 Qe at the ground line, as shown in Figure 12.24a. A superposition solution
can now be used to obtain the ground line deflection. The step-by-step procedure is as follows (refer to Figure 12.24b):
Step 1. Calculate Qg and Mg .
Step 2. Calculate the deflection xoQ that would be caused by the load Qg acting
alone.
Step 3. Calculate the deflection xoM that would be caused by the moment acting
alone.
Step 4. Determine the value of a load QgM that would cause the same deflection as
the moment (i.e., xoM).
Step 5. Determine the value of a moment MgQ that would cause the same deflection as the load (i.e., xoQ).
Step 6. Calculate (Qg 1 QgM )>Qc . and determine xoQM>Ds .
Step 7. Calculate (Mg 1 MgQ )>Mc and determine xoMQ>Ds .
12.12 Lateral Load-Carrying Capacity—Characteristic Load and Moment Method
Q
673
Mg Qe
e
Qg Q
xo
xoQM xoMQ
0.5
Ds
Ds
Ds
(
Q
)
(a)
Qg
Qc
Mg
Mc
Qg QgM
Qc
QgM
Qc
Qg
Qc
Step 6
Mg
Mc
Step 4
Step 7
Step 4
Qg
Qc
Step 7
Mg MgQ
Mc
Step 2
Step 3
Step 3
Step 5
Mg
Mc
MgQ
Mc
Step 2
xoQ
Ds
xoM
Ds
xo
Ds
xoQM
Ds
xoMQ
Ds
(b)
Figure 12.24 Superposition of deflection due to load and moment
Step 8. Calculate the combined deflection:
xo(combined) 5 0.5(xoQM 1 xoMQ )
(12.47)
Maximum Moment in Drilled Shaft Due to Ground Line Load Only
Figure 12.25 shows the plot of Qg>Qc with Mmax>Mc for fixed- and free-headed
drilled shafts due only to the application of a ground line load Qg . For fixed-headed
674 Chapter 12: Drilled-Shaft Foundations
0.045
0.020
Fixed
Free
0.015
Fixed
Free
0.030
Qg
Qc
0.010
Qg
Qc
(Sand)
(Clay)
0.015
Clay
0.005
Sand
0
0
0.005
0
0.015
0.010
Mmax
Mc
Figure 12.23 Variation of
Qg
Qc
with
Mmax
Mc
shafts, the maximum moment in the shaft, Mmax , occurs at the ground line. For this condition, if Qc , Mc , and Qg are known, the magnitude of Mmax can be easily calculated.
Maximum Moment Due to Load and Moment at Ground Line
If a load Qg and a moment Mg are applied at the ground line, the maximum moment in the
drilled shaft can be determined in the following manner:
Step 1. Using the procedure described before, calculate xo(combined) from
Eq. (12.47).
Step 2. To solve for the characteristic length T, use the following equation:
xo(combined) 5
2.43Qg
EpIp
T3 1
1.62Mg
EpIp
T2
(12.48)
Step 3. The moment in the shaft at a depth z below the ground surface can be calculated as
Mz 5 A mQgT 1 BmMg
(12.49)
12.12 Lateral Load-Carrying Capacity—Characteristic Load and Moment Method
675
where A m , Bm 5 dimensionless moment coefficients (Matlock and Reese,
1961); see Figure 12.26.
The value of the maximum moment Mmax can be obtained by calculating Mz at various depths in the upper part of the drilled shaft.
The characteristic load method just described is valid only if L>Ds has a certain minimum value. If the actual L>Ds is less than (L>Ds ) min , then the ground line deflections will
be underestimated and the moments will be overestimated. The values of (L>Ds ) min for
drilled shafts in sand and clay are given in the following table:
Clay
Sand
EpRI
EpRI
cu
(L/Ds)min
g9Dsf9Kp
(L/Ds)min
1 3 105
3 3 105
1 3 106
3 3 106
6
10
14
18
1 3 104
4 3 104
2 3 105
8
11
14
Am , Bm
0
0.2
0.4
0.6
0.8
1.0
0
Am
0.5
Bm
z
1.0
T
1.5
2.0
Figure 12.26 Variation of A m and Bm with z>T
68125_12_ch12_p637-685.qxd
3/12/10
11:38 AM
Page 676
676 Chapter 12: Drilled-Shaft Foundations
Example 12.7
A free-headed drilled shaft in clay is shown in Figure 12.27. Let Ep 5 22 3106 kN>m2.
Determine
a.
b.
c.
d.
The ground line deflection, xo(combined)
The maximum bending moment in the drilled shaft
The maximum tensile stress in the shaft
The minimum penetration of the shaft needed for this analysis
Solution
We are given
Ds 5 1 m
cu 5 100 kN>m2
RI 5 1
Ep 5 22 3 106 kN>m2
and
Ip 5
pD4s
(p) (1) 4
5
5 0.049 m4
64
64
Part a
From Eq. (12.43),
Qc 5 7.34D2s (EpRI ) ¢
cu 0.68
≤
EpRI
5 (7.34) (1) 2 3(22 3 106 ) (1)4 B
0.68
100
R
(22 3 106 ) (1)
5 37,607 kN
Mg 200 kN-m
Qg 150 kN
Clay
cu 100 kN/m2
Ds 1m
Figure 12.27 Free-headed drilled shaft
12.12 Lateral Load-Carrying Capacity—Characteristic Load and Moment Method
677
From Eq. (12.45),
Mc 5 3.86D3s (EpRI ) ¢
cu 0.46
≤
EpRI
5 (3.86) (1) 3 3(22 3 106 ) (1)4 B
0.46
100
R
6
(22 3 10 ) (1)
5 296,139 kN-m
Thus,
Qg
5
Qc
150
5 0.004
37,607
From Figure 12.22, xoQ < (0.0025)Ds 5 0.0025 m 5 2.5 mm. Also,
Mg
Mc
5
200
5 0.000675
296,139
From Figure 12.22, xoM < (0.0014)Ds 5 0.0014 m 5 1.4 mm, so
xoM
0.0014
5
5 0.0014
Ds
1
From Figure 12.22, for xoM>Ds 5 0.0014, the value of QgM>Qc < 0.002. Hence,
xoQ
Ds
5
0.0025
5 0.0025
1
From Figure 12.22, for xoQ>Ds 5 0.0025, the value of MgQ>Mc < 0.0013, so
Qg
Qc
1
QgM
Qc
5 0.004 1 0.002 5 0.006
From Figure 12.22, for (Qg 1 QgM )>Qc 5 0.006, the value of xoQM>Ds < 0.0046.
Hence,
xoQM 5 (0.0046) (1) 5 0.0046 m 5 4.6 mm
Thus, we have
Mg
Mc
1
MgQ
Mc
5 0.000675 1 0.0013 < 0.00198
From Figure 12.22, for (Mg 1 MgQ )>Mc 5 0.00198, the value of xoMQ>Ds < 0.0041.
Hence,
xoMQ 5 (0.0041) (1) 5 0.0041 m 5 4.1 mm
678 Chapter 12: Drilled-Shaft Foundations
Consequently,
xo (combined) 5 0.5(xoQM 1 xoMQ ) 5 (0.5) (4.6 1 4.1) 5 4.35 mm
Part b
From Eq. (12.48),
2.43Qg
xo (combined) 5
EpIp
T3 1
1.62Mg
EpIp
T2
so
(2.43) (150)
0.00435 m 5
6
(22 3 10 ) (0.049)
T3 1
(1.62) (200)
(22 3 106 ) (0.049)
T2
or
0.00435 m 5 338 3 1026 T3 1 300.6 3 1026 T2
and it follows that
T < 2.05 m
From Eq. (12.49),
Mz 5 A m Qg T 1 Bm Mg 5 A m (150) (2.05) 1 Bm (200) 5 307.5A m 1 200 Bm
Now the following table can be prepared:
z
T
Am
(Figure 12.26)
0
0.4
0.6
0.8
1.0
1.1
1.25
Bm
(Figure 12.26)
0
0.36
0.52
0.63
0.75
0.765
0.75
Mz (kN-m)
1.0
0.98
0.95
0.9
0.845
0.8
0.73
200
306.7
349.9
373.7
399.6
395.2
376.6
So the maximum moment is 399.4 kN-m < 400 kN-m and occurs at z>T < 1. Hence,
z 5 (1) (T) 5 (1) (2.05 m) 5 2.05 m
Part c
The maximum tensile stress is
Mmax ¢
s tensile 5
Ip
Ds
≤
2
5
1
(400) ¢ ≤
2
0.049
5 4081.6 kN , m2
12.13 Drilled Shafts Extending into Rock
679
Part d
We have
EpRI
cu
5
(22 3 106 ) (1)
5 2.2 3 105
100
By interpolation, for (EpRI )>cu 5 2.2 3 105, the value of (L>D s ) min < 8.5. So
L < (8.5) (1) 5 8.5 m
12.13
■
Drilled Shafts Extending into Rock
In Section 12.1, we noted that drilled shafts can be extended into rock. Figure 12.28
shows a drilled shaft whose depth of embedment in rock is equal to L. When considering drilled shafts in rock, we can find several correlations between the end bearing
capacity and the unconfined compression strength of intact rocks, qu. It is important to
recognize that, in the field, there are cracks, joints, and discontinuities in the rock, and
the influence of those factors should be considered. Keeping this in mind, Zhang and
Einstein (1998) analyzed a data base of 39 full-scale drilled shaft tests in which the
Qu
Soil
Rock
z
f
L
f unit side
resistance
qp unit point
bearing
Ds Db
qp
Figure 12.28 Drilled
shaft socketed into rock
680 Chapter 12: Drilled-Shaft Foundations
shaft bases were cast on or in generally soft rock with some degree of jointing. Based
on these results, they proposed
Qu(net) 5 Qp 1 Qs 5 qpA p 1 fpL
(12.50)
where end bearing capacity Qp can be expressed as
Qp (MN) 5 qpA p 5 34.83(qu MN>m2 ) 0.5143A p (m2 )4
(12.51)
Figure 12.29 shows the plot of qp (MN>m2 ) versus qu (MN>m2 ) obtained from the data on
which Eq. (12.51) is based.
Also, the side resistance Qs is
Qs (MN) 5 fpL 5 30.4(qu MN>m2 ) 0.543pDs (m)43L(m) 4
(for smooth socket)
(12.52)
Qs (MN) 5 fpL 5 30.8(qu MN>m2 ) 0.543pDs (m)43L(m) 4
(for rough socket)
(12.53)
and
100
qp(MN/m2)
30
10
qp = 4.83 (qu)0.51
3
1
0.1
30
0.3
1
3
10
Unconfined compression strength, qu (MN/m2)
Figure 12.29 Plot of qp versus qu (Adapted from Zhang and
Einstein, 1998)
100
Problems
681
Example 12.8
Figure 12.30 shows a drilled shaft extending into a shale formation. For the intact
rock cores, given qu 5 4.2 MN>m2. Estimate the allowable load-bearing capacity of
the drilled shaft. Use a factor of safety (FS) 5 3. Assume a smooth socket for side
resistance.
Solution
From Eq. (12.51),
Qp 5 A p 34.83(qu ) 0.514 5
p
(1) 2 3(4.83) (4.2) 0.514 5 7.89 MN
4
Again, from Eq. (12.52),
Qs 5 0.4(qu ) 0.5 (pDsL) 5 0.4(4.2) 0.5 3(p) (1) (4)4 5 10.3 MN
Hence,
Qall 5
3m
Qp 1 Qs
Qu
7.89 1 10.3
5
5
5 6.06MN
FS
FS
3
Soft clay
Ds 1 m
Shale
Smooth socket
4m
Drilled shaft
Figure 12.30 Drilled shaft extending into rock
■
Problems
12.1 A drilled shaft is shown in Figure P12.1. Determine the net allowable point bearing
capacity. Given
gc 5 15.6 kN>m3
Db 5 2 m
Ds 5 1.2 m
gs 5 17.6 kN>m3
L1 5 6 m
fr 5 35°
L2 5 3 m
cu 5 35 kN>m2
Factor of safety 5 3
Use Eq. (12.18).
12.2 Redo Problem 12.1, this time using Eq. (12.15). Let Es 5 600pa .
12.3 For the drilled shaft described in Problem 12.1, what skin resistance would
develop in the top 6 m, which are in clay? Use Eqs. (12.36) and (12.38).
682 Chapter 12: Drilled-Shaft Foundations
L1
Silty clay
Ds
␥c
cu
Sand
␥s
␾
c 0
L2
Db
Figure P12.1
12.4 Redo Problem 12.1 with the following:
Db 5 1.75 m
gc 5 17.8 kN>m3
Ds 5 1 m
gs 5 18.2 kN>m3
L1 5 6.25 m
fr 5 32°
L2 5 2.5 m
cu 5 32 kN>m2
Factor of safety 5 4
12.5 Redo Problem 12.4 using Eq. (12.5). Let Es 5 400pa .
12.6 For the drilled shaft described in Problem 12.4, what skin friction would develop
in the top 6.25 m?
a. Use Eqs. (12.36) and (12.38).
b. Use Eq. (12.39).
12.7 Figure P12.7 shows a drilled shaft without a bell. Assume the following values:
cu(1) 5 50 kN>m2
L1 5 6 m
L2 5 7 m
cu(2) 5 75 kN>m2
Ds 5 1.5 m
Determine:
a. The net ultimate point bearing capacity [use Eqs. (12.33) and (12.34)]
b. The ultimate skin friction [use Eqs. (12.36) and (12.38)]
c. The working load Qw (factor of safety 5 3)
12.8 Repeat Problem 12.7 with the following data:
L1 5 6.1 m cu(1) 5 70 kN>m2
L2 5 3.05 m cu(2) 5 120 kN>m2
Ds 5 0.91 m
Use Eqs. (12.39) and (12.40).
12.9 A drilled shaft in a medium sand is shown in Figure P12.9. Using the method proposed by Reese and O’Neill, determine the following:
Problems
L1
683
Clay
Ds
cu(1)
Clay
L2
cu(2)
Figure P12.7
a. The net allowable point resistance for a base movement of 25 mm
b. The shaft frictional resistance for a base movement of 25 mm
c. The total load that can be carried by the drilled shaft for a total base movement
of 25 mm
Assume the following values:
g 5 18 kN>m3
fr 5 38°
Dr 5 65%(medium sand)
L 5 12 m
L1 5 11 m
Ds 5 1 m
Db 5 2 m
Ds
Medium sand
L1
L
␥
␾
Average standard
penetration number (N60)
within 2Db below the
drilled shaft 19
Db
Figure P12.9
684 Chapter 12: Drilled-Shaft Foundations
12.10 In Figure P12.9, let L 5 7 m, L1 5 6 m, Ds 5 0.75 m, Db 5 1.25 m,
g 5 18 kN>m3, and fr 5 37°. The average uncorrected standard penetration number (N60 ) within 2Db below the drilled shaft is 29. Determine
a. The ultimate load-carrying capacity
b. The load-carrying capacity for a settlement of 12 mm.
The sand has 35% gravel. Use Eq. (12.30) and Figures 12.12 and 12.14.
12.11 For the drilled shaft described in Problem 12.7, determine
a. The ultimate load-carrying capacity
b. The load carrying capacity for a settlement of 25 mm
Use the procedure outlined by Reese and O’Neill. (See Figures 12.18 and 12.19.)
12.12 For the drilled shaft described in Problem 12.7, estimate the total elastic settlement
at working load. Use Eqs. (11.73), (11.75), and (11.76). Assume that
Ep 5 20 3 106 kN>m2, Cp 5 0.03, j 5 0.65, ms 5 0.3, Es 5 12,000 kN>m2, and
Q ws 5 0.8Qw . Use the value of Qw from Part (c) of Problem 12.7.
12.13 For the drilled shaft described in Problem 12.8, estimate the total elastic settlement
at working load. Use Eqs. (11.73), (11.75), and (11.76). Assume that
Ep 5 3 3 106 lb>in2, Cp 5 0.03, j 5 0.65, ms 5 0.3, Es 5 2000 lb>in2, and
Q ws 5 0.83Qw . Use the value of Qw from Part (c) of Problem 12.8.
12.14 Figure P12.14 shows a drilled shaft extending into clay shale. Given:
qu (clay shale) 5 1.81 MN>m2. Considering the socket to be rough, estimate the
allowable load-carrying capacity of the drilled shaft. Use FS 5 4.
12.15 A free-headed drilled shaft is shown in Figure P12.15. Let Qg 5 260 kN, Mg 5 0,
g 5 17.5 kN>m3, fr 5 35°, cr 5 0, and Ep 5 22 3 106 kN>m2. Determine
a. The ground line deflection, xo
b. The maximum bending moment in the drilled shaft
c. The maximum tensile stress in the shaft
d. The minimum penetration of the shaft needed for this analysis
Mg
2m
1.5 m
Loose sand
␥ 15 kN/m3
␾ 30˚
Qg
Clay shale
8m
Figure P12.14
Concrete
drilled shaft
Ds 1.25 m
Figure P12.15
␥
c, cu
␾, ␾
References
685
References
BEREZANTZEV, V. G., KHRISTOFOROV, V. S., and GOLUBKOV, V. N. (1961). “Load Bearing Capacity and
Deformation of Piled Foundations,” Proceedings, Fifth International Conference on Soil Mechanics and Foundation Engineering, Paris, Vol. 2, pp. 11–15.
CHEN, Y.-J., and KULHAWY, F. H. (1994). “Case History Evaluation of the Behavior of Drilled Shafts
under Axial and Lateral Loading,” Final Report, Project 1493-04, EPRI TR-104601, Geotechnical Group, Cornell University, Ithaca, NY, December.
DUNCAN, J. M., EVANS, L. T., JR., and Ooi, P. S. K. (1994). “Lateral Load Analysis of Single Piles and
Drilled Shafts,” Journal of Geotechnical Engineering, ASCE, Vol. 120, No. 6, pp. 1018–1033.
KULHAWY, F. H., and JACKSON, C. S. (1989). “Some Observations on Undrained Side Resistance of
Drilled Shafts,” Proceedings, Foundation Engineering: Current Principles and Practices,
American Society of Civil Engineers, Vol. 2, pp. 1011–1025.
MATLOCK, H., and REESE, L.C. (1961). “Foundation Analysis of Offshore Pile-Supported Structures,” in Proceedings, Fifth International Conference on Soil Mechanics and Foundation
Engineering, Vol. 2, Paris, pp. 91–97.
O’NEILL, M. W. (1997). Personal communication.
O’NEILL, M.W., and REESE, L.C. (1999). Drilled Shafts: Construction Procedure and Design Methods, FHWA Report No. IF-99-025.
REESE, L. C., and O’NEILL, M. W. (1988). Drilled Shafts: Construction and Design, FHWA, Publication No. HI-88-042.
REESE, L. C., and O’NEILL, M. W. (1989). “New Design Method for Drilled Shafts from Common
Soil and Rock Tests,” Proceedings, Foundation Engineering: Current Principles and Practices, American Society of Civil Engineers, Vol. 2, pp. 1026–1039.
REESE, L. C., TOUMA, F. T., and O’NEILL, M. W. (1976). “Behavior of Drilled Piers under Axial
Loading,” Journal of Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 102, No. GT5, pp. 493–510.
ROLLINS, K. M., CLAYTON, R. J., MIKESELL, R. C., and BLAISE, B. C. (2005). “Drilled Shaft Side
Friction in Gravelly Soils,” Journal of Geotechnical and Geoenvironmental Engineering,
American Society of Civil Engineers, Vol. 131, No. 8, pp. 987–1003.
WHITAKER, T., and COOKE, R. W. (1966). “An Investigation of the Shaft and Base Resistance of Large
Bored Piles in London Clay,” Proceedings, Conference on Large Bored Piles, Institute of
Civil Engineers, London, pp. 7–49.
ZHANG, L., and EINSTEIN, H. H. (1998). “End Bearing Capacity of Drilled Shafts in Rock,”
Journal of Geotechnical and Geoenvironmental Engineering, American Society of Civil
Engineers, Vol. 124, No. 7, pp. 574–584.
68125_13_ch13_p686-721.qxd
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13.1
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Page 686
Foundations on Difficult Soils
Introduction
In many areas of the United States and other parts of the world, certain soils make the
construction of foundations extremely difficult. For example, expansive or collapsible
soils may cause high differential movements in structures through excessive heave
or settlement. Similar problems can also arise when foundations are constructed over
sanitary landfills. Foundation engineers must be able to identify difficult soils when
they are encountered in the field. Although not all the problems caused by all soils can
be solved, preventive measures can be taken to reduce the possibility of damage to
structures built on them. This chapter outlines the fundamental properties of three
major soil conditions—collapsible soils, expansive soils, and sanitary landfills—and
methods of careful construction of foundations.
Collapsible Soil
13.2
Definition and Types of Collapsible Soil
Collapsible soils, which are sometimes referred to as metastable soils, are unsaturated
soils that undergo a large change in volume upon saturation. The change may or may
not be the result of the application of additional load. The behavior of collapsing soils
under load is best explained by the typical void ratio effective pressure plot (e against
log sr ) for a collapsing soil, as shown in Figure 13.1. Branch ab is determined from the
consolidation test on a specimen at its natural moisture content. At an effective pressure level of swr , the equilibrium void ratio is e1 . However, if water is introduced into
the specimen for saturation, the soil structure will collapse. After saturation, the equilibrium void ratio at the same effective pressure level swr is e2 ; cd is the branch of the
e–log sr curve under additional load after saturation. Foundations that are constructed
on such soils may undergo large and sudden settlement if the soil under them becomes
saturated with an unanticipated supply of moisture. The moisture may come from any
of several sources, such as (a) broken water pipelines, (b) leaky sewers, (c) drainage
686
13.3 Physical Parameters for Identification
e1
Void ratio, e
e2
Effective
pressure,
␴ (log scale)
␴w
a
b
687
Water added
c
d
Figure 13.1 Nature of variation of void ratio
with pressure for a collapsing soil
from reservoirs and swimming pools, (d) a slow increase in groundwater, and so on.
This type of settlement generally causes considerable structural damage. Hence, identifying collapsing soils during field exploration is crucial.
The majority of naturally occurring collapsing soils are aeolian—that is, winddeposited sands or silts, such as loess, aeolic beaches, and volcanic dust deposits. The
deposits have high void ratios and low unit weights and are cohesionless or only slightly
cohesive. Loess deposits have silt-sized particles. The cohesion in loess may be the result
of clay coatings surrounding the silt-size particles. The coatings hold the particles in a
rather stable condition in an unsaturated state. The cohesion also may be caused by the
presence of chemical precipitates leached by rainwater. When the soil becomes saturated,
the clay binders lose their strength and undergo a structural collapse. In the United States,
large parts of the Midwest and the arid West have such types of deposit. Loess deposits are
also found over 15 to 20% of Europe and over large parts of China. The thickness of
loess deposits can range up to about 10 m in the central United States. In parts of China it
can be up to 100 m. Figure 13.2 shows the extent of the loess deposits in the Mississippi
River basin.
Many collapsing soils may be residual soils that are products of the weathering of
parent rocks. Weathering produces soils with a large range of particle-size distribution.
Soluble and colloidal materials are leached out by weathering, resulting in large void
ratios and thus unstable structures. Many parts of South Africa and Zimbabwe have
residual soils that are decomposed granites. Sometimes collapsing soil deposits may be
left by flash floods and mudflows. These deposits dry out and are poorly consolidated.
An excellent review of collapsing soils is that of Clemence and Finbarr (1981).
13.3
Physical Parameters for Identification
Several investigators have proposed various methods for evaluating the physical parameters of collapsing soils for identification. Some of these methods are discussed briefly in
Table 13.1.
Jennings and Knight (1975) suggested a procedure for describing the collapse
potential of a soil: An undisturbed soil specimen is taken at its natural moisture content in a consolidation ring. Step loads are applied to the specimen up to a pressure
level swr of 200 kN>m2. (In Figure 13.1, this is swr .) At that pressure, the specimen
688 Chapter 13: Foundations on Difficult Soils
Minnesota
Wisconsin
South Dakota
Iowa
Indiana
Nebraska
Illinois
Kansas
Missouri
Kentucky
Tennessee
Arkansas
Alabama
Mississippi
Louisiana
Figure 13.2 Loess deposit in Mississippi River basin
is flooded for saturation and left for 24 hours. This test provides the void ratios e1 and
e2 before and after flooding, respectively. The collapse potential may now be calculated as
Cp 5 De 5
e1 2 e2
1 1 eo
(13.1)
where
eo 5 natural void ratio of the soil
De 5 vertical strain
The severity of foundation problems associated with a collapsible soil have been correlated
with the collapse potential Cp by Jennings and Knight (1975). They were summarized by
Clemence and Finbarr (1981) and are given in Table 13.2.
Holtz and Hilf (1961) suggested that a loessial soil that has a void ratio large enough to
allow its moisture content to exceed its liquid limit upon saturation is susceptible to collapse.
So, for collapse,
w(saturated) > LL
(13.2)
13.3 Physical Parameters for Identification
689
where LL 5 liquid limit.
However, for saturated soils,
eo 5 wGs
(13.3)
where Gs 5 specific gravity of soil solids.
Table 13.1 Reported Criteria for Identification of Collapsing Soila
a
Investigator
Year
Criteria
Denisov
1951
Clevenger
1958
Priklonski
1952
Gibbs
1961
Soviet Building Code
1962
Feda
1964
Benites
1968
Handy
1973
Coefficient of subsidence:
void ratio at liquid limit
K5
natural void ratio
K 5 0.5– 0.75: highly collapsible
K 5 1.0: noncollapsible loam
K 5 1.5 –2.0: noncollapsible soils
If dry unit weight is less than
12.6 kN>m3, settlement will be large;
if dry unit weight is greater than
14 kN>m3 settlement will be small.
natural moisture content 2 plastic limit
KD 5
plasticity index
KD , 0: highly collapsible soils
KD . 0.5: noncollapsible soils
KD . 1.0: swelling soils
saturation moisture content
Collapse ratio, R 5
liquid limit
This was put into graph form.
eo 2 eL
L5
1 1 eo
where eo 5 natural void ratio and eL 5 void ratio at
liquid limit. For natural degree of saturation less than
60%, if L . 20.1, the soil is a collapsing soil.
wo
PL
2
KL 5
Sr
PI
where wo 5 natural water content, Sr 5 natural degree
of saturation, PL 5 plastic limit, and PI 5 plasticity
index. For Sr , 100%, if KL . 0.85, the soil is a
subsident soil.
A dispersion test in which 2 g of soil are dropped into
12 ml of distilled water and specimen is timed
until dispersed; dispersion times of 20 to 30 s were
obtained for collapsing Arizona soils.
Iowa loess with clay (,0.002 mm) contents:
,16%: high probability of collapse
16–24%: probability of collapse
24–32%: less than 50% probability of collapse
.32%: usually safe from collapse
Modified after Lutenegger and Saber (1988)
690 Chapter 13: Foundations on Difficult Soils
Table 13.2 Relation of Collapse Potential to the Severity of Foundation Problemsa
Cp (%)
Severity of problem
0–1
1–5
5–10
10–20
20
No problem
Moderate trouble
Trouble
Severe trouble
Very severe trouble
a
From Clemence, S. P., and Finbarr, A. O. (1981). “Design Considerations for Collapsible Soils,”
Journal of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 107,
GT3 pp. 305–317. With permission from ASCE.
Combining Eqs. (13.2) and (13.3) for collapsing soils yields
eo > (LL) (Gs )
(13.4)
The natural dry unit weight of the soil required for its collapse is
gd <
Gsgw
Gsgw
5
1 1 eo
1 1 (LL) (Gs )
(13.5)
For an average value of Gs 5 2.65, the limiting values of gd for various liquid limits may
now be calculated from Eq. (13.5).
Figure 13.3 shows a plot of the preceding limiting values of dry unit weights against
the corresponding liquid limits. For any soil, if the natural dry unit weight falls below the
limiting line, the soil is likely to collapse.
20
Natural dry unit weight, ␥d (kN/m3)
18
16
14
Loessial soil
likely to
collapse
12
10
10
20
30
Liquid limit
40
45
Figure 13.3 Loessial soil likely
to collapse
13.4 Procedure for Calculating Collapse Settlement
691
Care should be taken to obtain undisturbed samples for determining the collapse
potentials and dry unit weights—preferably block samples cut by hand. The reason is that
samples obtained by thin-walled tubes may undergo some compression during the
sampling process. However, if cut block samples are used, the boreholes should be made
without water.
13.4
Procedure for Calculating Collapse Settlement
Jennings and Knight (1975) proposed the following laboratory procedure for determining
the collapse settlement of structures upon saturation of soil:
Step 1. Obtain two undisturbed soil specimens for tests in a standard consolidation
test apparatus (oedometer).
Step 2. Place the two specimens under 1 kN>m2 pressure for 24 hours.
Step 3. After 24 hours, saturate one specimen by flooding. Keep the other specimen at its natural moisture content.
Step 4. After 24 hours of flooding, resume the consolidation test on both specimens
by doubling the load (the same as in the standard consolidation test) to the
desired pressure level.
Step 5. Plot the e–log sr graphs for both specimens (Figures 13.4a and b).
Step 6. Calculate the in situ effective pressure, sor . Draw a vertical line corresponding to the pressure sor .
Step 7. From the e–log sor curve of the soaked specimen, determine the preconsolidation pressure, scr . If scr >sor 5 0.8–1.5, the soil is normally consolidated;
however, if scr >sor . 1.5, the soil is preconsolidated.
␴o
␴c ␴o ␴
␴o ,
eo
Specimen at
natural
moisture
content
e1
␴c ␴o ␴
␴o , eo
e1
e2
Soaked
specimen
Void ratio, e
␴o
log ␴
(a)
log ␴
Specimen at
natural
moisture
content
e2
Void ratio, e
Soaked
specimen
(b)
Figure 13.4 Settlement calculation from double oedometer test: (a) normally consolidated soil;
(b) overconsolidated soil
692 Chapter 13: Foundations on Difficult Soils
Step 8. Determine eor , corresponding to sor from the e–log sor curve of the soaked
specimen. (This procedure for normally consolidated and overconsolidated
soils is shown in Figures 13.4a and b, respectively.)
Step 9. Through point (sor , eor ) draw a curve that is similar to the e–log sor curve
obtained from the specimen tested at its natural moisture content.
Step 10. Determine the incremental pressure, Dsr, on the soil caused by the construction of the foundation. Draw a vertical line corresponding to the pressure of
sor 1 Dsr in the e–log sr curve.
Step 11. Now, determine De1 and De2 . The settlement of soil without change in the
natural moisture content is
Sc(1) 5
De1
(H)
1 1 eor
(13.6)
where H 5 thickness of soil susceptible to collapse.
Also, the settlement caused by collapse in the soil structure is
Sc(2) 5
(13.7)
Foundation Design in Soils Not Susceptible to Wetting
For actual foundation design purposes, some standard field load tests may also be conducted. Figure 13.5 shows the relation of the nature of load per unit arca versus settlement in a field load test in a loessial deposit. Note that the load–settlement relationship
is essentially linear up to a certain critical pressure, scrr , at which there is a breakdown
of the soil structure and hence a large settlement. Sudden breakdown of soil structure
is more common with soils having a high natural moisture content than with normally
dry soils.
If enough precautions are taken in the field to prevent moisture from increasing
under structures, spread foundations and mat foundations may be built on collapsible
Load per unit area
␴cr
Settlement
13.5
De2
(H)
1 1 eor
Figure 13.5 Field load test in loessial soil: load
per unit area versus settlement
13.5 Foundation Design in Soils Not Susceptible to Wetting
693
soils. However, the foundations must be proportioned so that the critical stresses (see
Figure 13.5) in the field are never exceeded. A factor of safety of about 2.5 to 3 should
be used to calculate the allowable soil pressure, or
sall
r 5
scrr
FS
(13.8)
where
sall
r 5 allowable soil pressure
FS 5 factor of safety
The differential and total settlements of these foundations should be similar to those of
foundations designed for sandy soils.
Continuous foundations may be safer than isolated foundations over collapsible soils
in that they can effectively minimize differential settlement. Figure 13.6 shows a typical
procedure for constructing continuous foundations. This procedure uses footing beams and
longitudinal load-bearing beams.
In the construction of heavy structures, such as grain elevators, over collapsible soils,
settlements up to about 0.3 m are sometimes allowed (Peck, Hanson, and Thornburn,
1974). In this case, tilting of the foundation is not likely to occur, because there is no
eccentric loading. The total expected settlement for such structures can be estimated from
standard consolidation tests on specimens of field moisture content. Without eccentric
loading, the foundations will exhibit uniform settlement over loessial deposits; however, if
the soil is of residual or colluvial nature, settlement may not be uniform. The reason is the
nonuniformity generally encountered in residual soils.
Extreme caution must be used in building heavy structures over collapsible soils. If
large settlements are expected, drilled-shaft and pile foundations should be considered.
These types of foundation can transfer the load to a stronger load-bearing stratum.
Load-bearing beams
Figure 13.6 Continuous foundation with load-bearing beams (From Clemence, S. P., and
Finbarr, A. O. (1981). “Design Considerations for Collapsible Soils,” Journal of the Geotechnical
Engineering Division, American Society of Civil Engineers, Vol. 107, GT3 pp. 305–317.
With permission from ASCE.)
694 Chapter 13: Foundations on Difficult Soils
Foundation Design in Soils Susceptible to Wetting
If the upper layer of soil is likely to get wet and collapse sometime after construction of
the foundation, several design techniques to avoid failure of the foundation may be considered. They are as follow:
Dynamic Compaction
If the expected depth of wetting is about 1.5 to 2 m from the ground surface, the soil may
be moistened and recompacted by heavy rollers. Spread footings and mats may be constructed over the compacted soil. An alternative to recompaction by heavy rollers is heavy
tamping, which is sometimes referred to as dynamic compaction. (See Chapter 14.) Heavy
tamping consists primarily of dropping a heavy weight repeatedly on the ground. The
height of the drop can vary from 8 to 30 m. The stress waves generated by the dropping
weight help in the densification of the soil.
Lutenegger (1986) reported the use of dynamic compaction to stabilize a thick layer of
friable loess before construction of a foundation in Russe, Bulgaria. During field exploration,
the water table was not encountered to a depth of 10 m, and the natural moisture content was
below the plastic limit. Initial density measurements made on undisturbed soil specimens indicated that the moisture content at saturation would exceed the liquid limit, a property usually
encountered in collapsible loess. For dynamic compaction of the soil, the upper 1.7 m of crust
material was excavated. A circular concrete weight of 133 kN was used as a hammer. At each
grid point, compaction was achieved by dropping the hammer 7 to 12 times through a vertical distance of 2.5 m.
Figure 13.7 shows the dry unit weight of the soil before and after compaction. The
increase in dry unit weight of the soil shows that dynamic compaction can be used effectively to stabilize collapsible soil.
Dry unit weght (kN/m3)
12
0
14
16
18
20
Before
compaction
1
After
compaction
2
Depth (m)
13.6
3
4
5
Figure 13.7 Dynamic compaction of a
friable loess layer in Russe, Bulgaria
(Adapted after Lutenegger, 1986)
13.7 General Nature of Expansive Soils
695
Chemical Stabilization
If conditions are favorable, foundation trenches can be flooded with solutions of sodium silicate and calcium chloride to stabilize the soil chemically. The soil will then behave like a soft
sandstone and resist collapse upon saturation. This method is successful only if the solutions
can penetrate to the desired depth; thus, it is most applicable to fine sand deposits. Silicates
are rather costly and are not generally used. However, in some parts of Denver, silicates have
been used successfully.
The injection of a sodium silicate solution to stabilize collapsible soil deposits has been
used extensively in the former Soviet Union and Bulgaria (Houston and Houston, 1989). This
process, which is used for dry collapsible soils and for wet collapsible soils that are likely to
compress under the added weight of the structure to be built, consists of three steps:
Step 1. Injection of carbon dioxide to remove any water that is present and for
preliminary activation of the soil
Step 2. Injection of sodium silicate grout
Step 3. Injection of carbon dioxide to neutralize alkalies.
Vibrofloation and Ponding
When the soil layer is susceptible to wetting to a depth of about 10 m (<30 ft), several
techniques may be used to cause collapse of the soil before construction of the foundation
is begun. Two of these techniques are vibroflotation and ponding (also called flooding).
Vibroflotation is used successfully in free-draining soil. (See Chapter 14.) Ponding—by
constructing low dikes—is utilized at sites that have no impervious layers. However, even
after saturation and collapse of the soil by ponding, some additional settlement of the soil
may occur after construction of the foundation is begun. Additional settlement may also be
caused by incomplete saturation of the soil at the time of construction. Ponding may be
used successfully in the construction of earth dams.
Extending Foundation Beyond Zone of Wetting
If precollapsing the soil is not practical, foundations may be extended beyond the zone of
possible wetting, although the technique may require drilled shafts and piles. The design
of drilled shafts and piles must take into consideration the effect of negative skin friction
resulting from the collapse of the soil structure and the associated settlement of the zone
of subsequent wetting.
In some cases, a rock-column type of foundation (vibroreplacement) may be considered. Rock columns are built with large boulders that penetrate the collapsible soil layer.
They act as piles in transferring the load to a more stable soil layer.
Expansive Soils
13.7
General Nature of Expansive Soils
Many plastic clays swell considerably when water is added to them and then shrink
with the loss of water. Foundations constructed on such clays are subjected to large
uplifting forces caused by the swelling. These forces induce heaving, cracking, and the
breakup of both building foundations and slab-on-grade members. Figure 13.8 shows
696 Chapter 13: Foundations on Difficult Soils
Figure 13.8 Cracks in a wall due to heaving of an expansive clay (Courtesy
of Anand Puppala, University of Texas at Arlington, Arlington, Texas)
the cracks in a wall due to excessive heaving. Expansive clays cover large parts of the
United States, South America, Africa, Australia, and India. In the United States, these clays
are predominant in Texas, Oklahoma, and the upper Missouri Valley. In general, expansive
clays have liquid limits and plasticity indices greater than about 40 and 15, respectively.
As noted, the increase and decrease in moisture content causes clay to swell and
shrink. Figure 13.9 shows shrinkage cracks on the ground surface of a clay weathered from
the Eagle Ford shale formation in the Dallas-Fort Worth, Texas area. The depth in a soil to
which periodic changes of moisture occur is usually referred to as the active zone (see
Figure 13.10). The depth of the active zone varies, depending on the location of the site.
Some typical active-zone depths in American cities are given in Table 13.3. In some clays
and clay shales in the western United States, the depth of the active zone can be as much
as 15 m. The active-zone depth can easily be determined by plotting the liquidity index
against the depth of the soil profile over several seasons. Figure 13.11 shows such a plot
for the Beaumont formation in the Houston area.
13.7 General Nature of Expansive Soils
Figure 13.9 Shrinkage cracks on ground surface in a clay weathered from Eagle Ford shale
formation in the Dallas-Fort Worth area (Courtesy of Thomas M. Petry, Missouri University
of Science and Technology, Rolla, Missouri)
Ground
surface
Moisture
content
Seasonal variation
of moisture content
Active zone
(Depth z)
um nt
libri
e
Equi ure cont
t
mois
Depth
Figure 13.10 Definition of active zone
697
698 Chapter 13: Foundations on Difficult Soils
Table 13.3 Typical Active-Zone Depths in
Some U.S. Citiesa
City
Depth of active zone
(m)
Houston
Dallas
San Antonio
Denver
1.5 to 3
2.1 to 4.6
3 to 9
3 to 4.6
a
After O’Neill and Poormoayed (1980)
(O’Neill, M. W., and Poormoayed, N.(1980).
“Methodology for Foundations on Expansive
Clays,” Journal of the Geotechnical
Engineering Division, American Society
of Civil Engineers, Vol. 106, No. GT12,
pp. 1345–1367. With permission from ASCE.)
Shrinkage cracks can extend deep into the active zone. Figure 13.12 shows interconnected shrinkage cracks extending from the ground surface into the active zone in an
expansive clay.
To study the magnitude of possible swell in a clay, simple laboratory oedometer tests
can be conducted on undisturbed specimens. Two common tests are the unrestrained swell
test and the swelling pressure test. They are described in the following sections.
1
Liquidity index
0
1
1
Depth (m)
2
Approximate
depth
of seasonal
change, 1.67 m
3
Range over
several
seasons
4
5
6
Figure 13.11 Active zone in Houston
area, Beaumont formation (O’Neill, M. W.
and Poormoayed, N. (1980).
“Methodology for Foundations on
Expansive Clays,” Journal of the
Geotechnical Engineering Division,
American Society of Civil Engineers,
Vol. 106, No. GT12, pp. 1345–1367.
With permission from ASCE.)
13.8 Unrestrained Swell Test
699
Figure 13.12 Interconnected shrinkage cracks extended from the ground surface into the
active zone (Courtesy of Thomas M. Petry, Missouri University of Science and Technology,
Rolla, Missouri)
13.8
Unrestrained Swell Test
In the unrestrained swell test, the specimen is placed in an oedometer under a small surcharge
of about 6.9 kN>m2 (1 lb>in2 ). Water is then added to the specimen, and the expansion of the
volume of the specimen (i.e., its height; the area of cross section is constant) is measured until
equilibrium is reached. The percentage of free swell may than be expressed as the ratio
sw(free) (%) 5
DH
(100)
H
(13.9)
where
sw(free) 5 free swell, as a percentage
DH 5 height of swell due to saturation
H 5 original height of the specimen
Vijayvergiya and Ghazzaly (1973) analyzed various soil test results obtained in
this manner and prepared a correlation chart of the free swell, liquid limit, and natural
moisture content, as shown in Figure 13.13. O’Neill and Poormoayed (1980) developed
a relationship for calculating the free surface swell from this chart:
DSF 5 0.0033Zsw(free)
(13.10)
700 Chapter 13: Foundations on Difficult Soils
20
Percent swell, s␻(free) (%)
10
60
50
40
1
Liquid
limit
70
0.1
0
40
10
20
30
Natural moisture content (%)
50
Figure 13.13 Relation between
percentage of free swell, liquid limit, and
natural moisture content (From
Vijayvergiya, V. N. and Ghazzaly, O. I.
(1973). “Prediction of Swelling Potential
of Natural Clays,” Proceedings,
Third International Research and
Engineering Conference on Expansive
Clays, pp. 227–234. With permission
from ASCE.)
where
DSF 5 free surface swell
Z 5 depth of active zone
sw(free) 5 free swell, as a percentage
13.9
Swelling Pressure Test
The swelling pressure can be determined from two different types of tests. They are
•
•
Conventional consolidation test
Constant volume test
The two methods are briefly described here.
Conventional Consolidation Test
In this type of test, the specimen is placed in a oedometer under a small surcharge of about
6.9 kN/m2. Water is added to the specimen, allowing it to swell and reach an equilibrium position after some time. Subsequently, loads are added in convenient steps, and the specimen is
consolidated. The plot of specimen deformation (␦) versus log sr is shown in Figure 13.14. The
␦ versus log sr plot crosses the horizontal line through the point of initial condition. The pressure corresponding to the point of intersection is the zero swell pressure, ssw
r .
13.9 Swelling Pressure Test
701
Specimen
deformation, d
Swelling due
to addition of
water
+ve
0
Consolidation
test
Initial
condition
–ve
ssw
Pressure, s
Figure 13.14 Zero swell pressure from
conventional consolidation test
Constant Volume Test
The constant volume test can be conducted by taking a specimen in a consolidation ring and
applying a pressure equal to the effective overburden pressure, sor , plus the approximate
anticipated surcharge caused by the foundation, ssr . Water is then added to the specimen. As
the specimen starts to swell, pressure is applied in small increments to prevent swelling.
Pressure is maintained until full swelling pressure is developed on the specimen, at which
time the total pressure is
ssw
r 5 sor 1 ssr 1 s1r
(13.11)
where
ssw
r 5 total pressure applied to prevent swelling, or zero swell pressure
s1r 5 additional pressure applied to prevent swelling after addition of water
Figure 13.15 shows the variation of the percentage of swell with effective pressure during a
swelling pressure test. (For more information on this type of test, see Sridharan et al., 1986.)
Swell, sw (%)
sw (1)
Unloading
0
␴o ␴s
␴sw
Effective pressure
Figure 13.15
Swelling pressure
test
702 Chapter 13: Foundations on Difficult Soils
A ssw
r of about 20 to 30 kN>m2 is considered to be low, and a ssw
r of
1500 to 2000 kN>m2 is considered to be high. After zero swell pressure is attained, the
soil specimen can be unloaded in steps to the level of the effective overburden pressure,
sor . The unloading process will cause the specimen to swell. The equilibrium swell for
each pressure level is also recorded. The variation of the swell, sw in percent, and the
applied pressure on the specimen will be like that shown in Figure 13.15.
The constant volume test can be used to determine the surface heave, DS, for a foundation (O’Neill and Poormoayed, 1980) as given by the formula
n
DS 5 a 3sw(1) (%)4 (Hi ) (0.01)
(13.12)
i51
where
sw(1) (%) 5 swell, in percent, for layer i under a pressure of sor 1 ssr (see Figure 13.15)
DHi 5 thickness of layer i
⬘
) obtained from the
It is important to point out that the zero swell pressure (␴sw
conventional consolidation test and the constant volume test may not be the same.
Table 13.4 summarizes some laboratory test results of Sridharan et al. (1986) to illustrate this point. It also was shown by Sridharan et al. (1986) that the zero swell
pressure is a function of the dry unit weight of soil, but not of the initial moisture
content (Figure 13.16).
Table 13.4 Comparison of Zero Swell Pressure Obtained from Conventional Consolidation Tests
and Constant Volume Tests—Summary of Test Results of Sridharan et al. (1986)
⬘ (kN/m2)
␴sw
Soil
Liquid
limit
Plasticity
index
Initial void
ratio, ei
Consolidation
test
Constant
volume test
BC-1
BC-4
BC-5
BC-7
BC-8
80
98
96
96
94
44
57
65
66
62
0.893
1.002
0.742
0.572
0.656
294.3
382.6
500.3
1275.3
147.2
186.4
251.8
304.1
372.8
68.7
Example 13.1
A soil profile has an active zone of expansive soil of 1.83 m. The liquid limit and the
average natural moisture content during the construction season are 50% and 20%,
respectively. Determine the free surface swell.
Zero swell pressure, ssw
(kN/m2)
13.9 Swelling Pressure Test
703
500
Symbol
400
300
Moisture
content
0
8.2
15.8
18.2
200
100
0
11
12
13
14
15
15.7
Dry unit weight (kN/m3)
Figure 13.16 Plot of zero swell pressure with the dry unit weight of soil
(After Sridharan et al., 1986. Copyright ASTM INTERNATIONAL.
Reprinted with permission.)
Solution
From Figure 13.13 for LL 5 50% and w 5 20%, sw(free) 5 3%. From Eq. (13.10),
DSF 5 0.0033Zsw(free)
Hence,
DSF 5 0.0033(1.83) (3) (12) 5 18.12 mm
■
Example 13.2
A soil profile’s active-zone depth is 3.5 m. If a foundation is to be placed 0.5 m below
the ground surface, what would be the estimated total swell? The following data were
obtained from laboratory tests:
Depth (m)
Swell under effective
overburden and estimated
foundation surcharge pressure,
sw(1)(%)
0.5
1
2
3
2
1.5
0.75
0.25
Solution
The values of sw(1) (%) are plotted with depth in Figure 13.17a. The area of this diagram
will be the total swell. The trapezoidal rule yields
DS 5
1
1
1
1
c (1) (0 1 0.5) 1 (1) (0.5 1 1.1) 1 (1) (1.1 1 2) d
100 2
2
2
5 0.026 m 5 26 mm
704 Chapter 13: Foundations on Difficult Soils
0
Ground surface
2 Swel1, sw(1) (%)
1
0.5
1.0
0.5
0
26
1.5
1.1
1.5
2.0
Total Swell,
DS (mm)
1.1 m
1.5
0.75
2.5
2.5
0.5
3.0
0.25
3.5
3.5
Depth (m)
Depth (m)
(a)
(b)
Figure 13.17 (a) Variation of sw(1) with depth; (b) variation of ⌬S with depth
■
Example 13.3
In Example 13.2, if the allowable total swell is 10 mm, what would be the undercut necessary to reduce the total swell?
Solution
Using the procedure outlined in Example 13.2, we calculate the total swell at various
depths below the foundation from Figure 13.17a as follows:
Total swell, ⌬S (mm)
Depth (m)
3.5
0
3
01 c
2.5
1.5
0.5
1
1
(0.5) (0.25) d
5 0.000625 m 5 0.625 mm
2
100
1
1
0.000625 1
c (0.5) (0.25 1 0.5) d 5 0.0025 m 5 2.5 mm
100 2
1
1
0.0025 1
c (1) (0.5 1 1.1) d 5 0.0105 m 5 10.5 mm
100 2
26 mm
Plotted in Figure 13.17b, these total settlements show that a total swell of 10 mm
corresponds to a depth of 1.6 m below the ground surface.
Hence, the undercut below the foundation is 1.6 ⫺ 0.5 ⫽ 1.1 m. This soil should be
excavated, replaced by nonswelling soil, and recompacted.
■
13.10 Classification of Expansive Soil on the Basis of Index Tests
Classification of Expansive Soil
on the Basis of Index Tests
Classification systems for expansive soils are based on the problems they create
in the construction of foundations (potential swell). Most of the classifications
contained in the literature are summarized in Figure 13.18 and Table 13.5.
However, the classification system developed by the U.S. Army Waterways
Experiment Station (Snethen et al., 1977) is the one most widely used in the
United States. It has also been summarized by O’Neill and Poormoayed (1980);
see Table 13.6. Sridharan (2005) proposed an index called the free swell ratio to predict the clay type, potential swell classification, and dominant clay minerals present
3
Very High
2
High
1
Medium
Low
0
Swelling
Potential
25%
5%
1.5%
Plasticity index (%)
Activity
100
80
n
U
40
Li
0
Suction(pF)
Very High
0
50
100
Percent of clay (2 ␮) in whole sample
(c)
0.7
L
20 40 60 80 100 120 140 160
Liquid limit (%)
(b)
7
Low
A
ine
)
0
I
II
III
IV
V
6
0
(
0.
20
L-
L
3(
20
100
High
Medium
e
Extra high
)
-8
LL
9
60
0 10 20 30 40 50 60 70 80 90 100
Percent clay sizes (finer than 0.002 mm)
(a)
50
Very
high
High
120
Med
4
Swelling
Low
Nonplastic
5
Plasticity index of whole sample
13.10
705
5
Special case
High
Moderate
Low
Nonexpansive
4
I
3
V
IV
III
II
2
1
0
10
20
30
40
50
Soil water content
(d)
60
Figure 13.18 Commonly used criteria for determining swell potential (From Abduljauwad, S. N.
and Al-Sulaimani, G. J. (1993). “Determination of Swell Potential of Al-Qatif Clay,” Geotechnical
Testing Journal, American Society for Testing and Materials, vol. 16, No. 4, pp. 469–484.
Copyright ASTM INTERNATIONAL. Reprinted with permission.)
Table 13.5 Summary of Some Criteria for Identifying Swell Potential (From Abduljauwad, S. N. and Al-Sulaimani,
G. J. (1993). “Determination of Swell Potential of Al-Qatif Clay,” Geotechnical Testing Journal, American Society for
Testing and Materials, vol. 16, No. 4, pp. 469–484. Copyright ASTM INTERNATIONAL. Reprinted with permission.)
Reference
Criteria
Remarks
Holtz (1959)
CC . 28, PI . 35, and SL , 11 (very high)
20 < CC < 31, 25 < PI < 41, and 7 < SL < 12 (high)
13 < CC < 23, 15 < PI < 28, and 10 < SL < 16
(medium)
CC < 15, PI < 18, and SL > 15 (low)
Based on CC, PI, and SL
Seed et al. (1962)
See Figure 13.18a
Based on oedometer test using
compacted specimen, percentage of clay ,2 mm, and activity
Altmeyer (1955)
LS , 5, SI . 12, and PS , 0.5 (noncritical)
5 < LS < 8, 10 < SL < 12, and 0.5 < PS < 1.5
(marginal)
LS . 8, SL , 10, and PS . 1.5 (critical)
Based on LS, SL, and PS
Remolded sample (rd(max) and
wopt) soaked under 6.9 kPa
surcharge
Dakshanamanthy
See Figure 13.18b
and Raman (1973)
Based on plasticity chart
Raman (1967)
PI . 32 and SI . 40 (very high)
23 < PI < 32 and 30 < SI < 40 (high)
12 < PI < 23 and 15 < SI < 30 (medium)
PI , 12 and SI , 15 (low)
Based on PI and SI
Sowers and Sowers
(1970)
SL , 10 and PI . 30 (high)
10 < SL < 12 and 15 < PI < 30 (moderate)
SL . 12 and PI , 15 (low)
Little swell will occur when wo
results in LI of 0.25
Van Der Merwe
(1964)
See Figure 13.18c
Based on PI, percentage of clay
,2 mm, and activity
Uniform Building
Code, 1968
EI . 130 (very high) and 91 < EI < 130 (high)
51 < EI < 90 (medium) and 21 < EI < 50 (low)
0 < EI < 20 (very low)
Based on oedometer test on compacted specimen with degree
of saturation close to 50% and
surcharge of 6.9 kPa
Snethen (1984)
LL . 60, PI . 35, tnat . 4, and SP . 1.5 (high)
30 < LL < 60, 25 < PI < 35, 1.5 < tnat < 4,
and 0.5 < SP < 1.5 (medium)
LL , 30, PI , 25, tnat , 1.5, and SP , 0.5 (low)
PS is representative for field
condition and can be used
without tnat , but accuracy will
be reduced
Chen (1988)
PI $ 35 (very high) and 20 < PI < 55 (high)
10 < PI < 35 (medium) and PI < 15 (low)
Based on PI
McKeen (1992)
Figure 13.18d
Based on measurements of
soil water content, suction, and
change in volume on drying
Vijayvergiya and
Ghazzaly (1973)
log SP 5 (1>12) (0.44 LL 2 wo 1 5.5)
Empirical equations
Nayak and Christensen (1974)
SP 5 (0.00229 PI) (1.45C)>wo 1 6.38
Empirical equations
Weston (1980)
SP 5 0.00411(LL w ) 4.17q23.86w22.33
o
Empirical equations
706
13.10 Classification of Expansive Soil on the Basis of Index Tests
707
Table 13.5 (continued)
Note: C 5 clay, %
CC 5 colloidal content, %
EI 5 Expansion index 5 100 3 percent swell 3 fraction
passing No. 4 sieve
LI 5 liquidity index, %
LL 5 liquid limit, %
LL w 5 weighted liquid limit, %
LS 5 linear shrinkage, %
PI 5 plasticity index, %
PS 5 probable swell, %
q 5 surcharge
SI 5 shrinkage index 5 LL 2 SL, %
SL 5 shrinkage limit, %
SP 5 swell potential, %
wo 5 natural soil moisture
wopt 5 optimum moisture content, %
tnat 5 natural soil suction in tsf
rd(max) 5 maximum dry density
Table 13.6 Expansive Soil Classification Systema
Liquid
limit
Plasticity
index
Potential
swell (%)
Potential swell
classification
Low
,50
,25
,0.5
50–60
25–35
0.5–1.5
Marginal
High
.60
.35
.1.5
Potential swell 5 vertical swell under a pressure equal to
overburden pressure
a
Compiled from O’Neill and Poormoayed (1980)
in a given soil. The free swell ratio can be determined by finding the equilibrium
sediment volumes of 10 grams of an oven-dried specimen passing No. 40 U.S. sieve
(0.425 mm opening) in distilled water (Vd) and in CCl4 or kerosene (VK). The free
swell ratio (FSR) is defined as
FSR 5
Vd
VK
(13.13)
Table 13.7 gives the expansive soil classification based on free swell ratio. Also,
Figure 13.19 shows the classification of soil based on the free swell ratio.
Table 13.7 Expansive Soil Classification Based on Free Swell Ratio
Free
swell ratio
< 1.0
1.0–1.5
1.5–2.0
2.0–4.0
. 4.0
Clay type
Non-swelling
Mixture of swelling
and non-swelling
Swelling
Swelling
Swelling
Potential swell
classification
Dominant clay
mineral
Negligible
Low
Kaolinite
Kaolinite and
montmorillonite
Montmorillonite
Montmorillonite
Montmorillonite
Moderate
High
Very High
708 Chapter 13: Foundations on Difficult Soils
80
4
2
1
1
60
III C
III B
III A
I Kaolinitic soils
V d (cm3)
1.5
II (Kaolinitic + montmorillonitic) soils
1
III Montmorillonitic soils
40
A Moderately swelling
1
II
B Highly swelling
1
C Very highly swelling
20
I
0
0
20
VK (cm3)
40
Figure 13.19 Classification based on free swell ratio (Adapted from Sridharan, 2005)
13.11
Foundation Considerations for Expansive Soils
If a soil has a low swell potential, standard construction practices may be followed.
However, if the soil possesses a marginal or high swell potential, precautions need to be
taken, which may entail
1. Replacing the expansive soil under the foundation
2. Changing the nature of the expansive soil by compaction control, prewetting, installation of moisture barriers, or chemical stabilization
3. Strengthening the structures to withstand heave, constructing structures that are flexible enough to withstand the differential soil heave without failure, or constructing
isolated deep foundations below the depth of the active zone
One particular method may not be sufficient in all situations. Combining several techniques
may be necessary, and local construction experience should always be considered.
Following are some details regarding the commonly used techniques for dealing with
expansive soils.
13.11 Foundation Considerations for Expansive Soils
709
Replacement of Expansive Soil
When shallow, moderately expansive soils are present at the surface, they can be removed
and replaced by less expansive soils and then compacted properly.
Changing the Nature of Expansive Soil
1. Compaction: The heave of expansive soils decreases substantially when the soil is
compacted to a lower unit weight on the high side of the optimum moisture content
(possibly 3 to 4% above the optimum moisture content). Even under such conditions, a slab-on-ground type of construction should not be considered when the total
probable heave is expected to be about 38 mm or more.
2. Prewetting: One technique for increasing the moisture content of the soil is ponding
and hence achieving most of the heave before construction. However, this technique
may be time consuming because the seepage of water through highly plastic clays is
slow. After ponding, 4 to 5% of hydrated lime may be added to the top layer of the
soil to make it less plastic and more workable (Gromko, 1974).
3. Installation of moisture barriers: The long-term effect of the differential heave can be
reduced by controlling the moisture variation in the soil. This is achieved by providing
vertical moisture barriers about 1.5 m deep around the perimeter of slabs for the slab-ongrade type of construction. These moisture barriers may be constructed in trenches filled
with gravel, lean concrete, or impervious membranes.
4. Stabilization of soil: Chemical stabilization with the aid of lime and cement has
often proved useful. A mix containing about 5% lime is sufficient in most cases.
The effect of lime in stabilizing expansive soils, thereby reducing the shrinking
and swelling characteristics, can be demonstrated with the aid of Figure 13.20. For
this, expansive clay weathered from the Eagle Ford shale formation in the DallasFort Worth, Texas area was taken. Some of it was mixed with water to about its
liquid limit. It was placed in two molds that were about 152 mm long and 12.7 mm
⫻ 12.7 mm in cross section. Figure 13.20a shows the shrinkage of the soil specimens
in the mold in a dry condition. The same soil also was mixed with 6% lime (by dry
weight) and then with a similar amount of water and placed in six similar molds.
Figures 13.20b shows the shrinkage of the lime-stabilized specimens in a dry condition, which was practically negligible compared to that seen in Figure 13.20a. Lime
or cement and water are mixed with the top layer of soil and compacted. The addition
(a)
Figure 13.20 Shrinkage of expansive clay (Eagle Ford
soil) mixed with water to about its liquid limit in molds
of 152 mm ⫻ 12.7 mm ⫻ 12.7 mm: (a) without addition
of lime;
710 Chapter 13: Foundations on Difficult Soils
Figure 13.20 (continued) (b) with
addition of 6% lime by weight
(Courtesy of Thomas M. Petry,
Missouri University of Science and
Technology, Rolla, Missouri)
(b)
of lime or cement will decrease the liquid limit, the plasticity index, and the swell
characteristics of the soil. This type of stabilization work can be done to a depth of
1 to 1.5 m. Hydrated high-calcium lime and dolomite lime are generally used for
lime stabilization.
Another method of stabilization of expansive soil is the pressure injection of lime slurry
or lime–fly-ash slurry into the soil, usually to a depth of 4 to 5 m or and occasionally deeper
to cover the active zone. Further details of the pressure injection technique are presented in
Chapter 14. Depending on the soil conditions at a site, single or multiple injections can be
planned, as shown in Figure 13.21. Figure 13.22 shows the slurry pressure injection work for
a building pad. The stakes that are marked are the planned injection points. Figure 13.23
shows lime–fly-ash stabilization by pressure injection of the bank of a canal that had experienced sloughs and slides.
Plan
Section
Single injection
Double injection
Figure 13.21 Multiple lime
slurry injection planning for a
building pad
13.12 Construction on Expansive Soils
711
Figure 13.22 Pressure injection of lime slurry for a building pad (Courtesy of Hayward Baker Inc.,
Odenton, Maryland.)
Figure 13.23 Slope stabilization of a canal bank by pressure injection of lime–fly-ash
slurry (Courtesy of Hayward Baker Inc., Odenton, Maryland.)
13.12
Construction on Expansive Soils
Care must be exercised in choosing the type of foundation to be used on expansive soils.
Table 13.8 shows some recommended construction procedures based on the total predicted
heave, DS, and the length-to-height ratio of the wall panels. For example, the table proposes the use of waffle slabs as an alternative in designing rigid buildings that are capable
of tolerating movement. Figure 13.24 shows a schematic diagram of a waffle slab. In this
type of construction, the ribs hold the structural load. The waffle voids allow the expansion of soil.
712 Chapter 13: Foundations on Difficult Soils
Table 13.8 Construction Procedures for Expansive Clay Soilsa
Total predicted heave
(mm)
L, H 5 1.25
L, H 5 2.5
0 to 6.35
6.35 to 12.7
12.7
12.7 to 50.8
12.7 to 50.8
50.8 to 101.6
Recommended
construction
Method
Remarks
No precaution
Rigid building
tolerating movement
(steel reinforcement
as necessary)
Foundations:
Pads
Strip footings
mat (waffle)
Footings should be small and deep,
consistent with the soilbearing capacity.
Mats should resist bending.
Floor slabs:
Waffle
Tile
Slabs should be designed to resist
bending and should be
independent of grade beams.
Walls:
Walls on a mat should be as
flexible as the mat. There should
be no vertical rigid connections.
Brickwork should be strengthened
with tie bars or bands.
Joints:
Clear
Flexible
Contacts between structural
units should be avoided, or
flexible, waterproof material may
be inserted in the joints.
Building damping
movement
Walls:
Walls or rectangular building
Flexible
units should heave as a unit.
Unit construction
Steel frame
Foundations:
Three point
Cellular
Jacks
.50.8
.101.6
Building independent
of movement
Cellular foundations allow slight
soil expansion to reduce swelling
pressure. Adjustable jacks can be
inconvenient to owners. Threepoint loading allows motion
without duress.
Foundation drilled Smallest-diameter and widely
shaft:
spaced shafts compatible with the
Straight shaft
load should be placed.
Bell bottom
Clearance should be allowed
under grade beams.
Suspended floor:
Floor should be suspended on
grade beams 305 to 460 mm
above the soil.
a
Gromko, G. J., (1974). “Review of Expansive Soils,” Journal of the Geotechnical Engineering Division,
American Society of Civil Engineers, Vol. 100, No. GT6, pp. 667–687. With permission from ASCE.
13.12 Construction on Expansive Soils
Void
713
Void
Figure 13.24 Waffle slab
Dead load, D
Grade beam
Ground surface
z
Ds
U
Active
zone, Z
Drilled shafts
with bells
(a)
Db
(b)
Figure 13.25 (a) Construction of drilled shafts with bells and grade beam; (b) definition of
parameters in Eq. (13.14)
Table 13.8 also suggests the use of a drilled shaft foundation with a suspended floor slab
when structures are constructed independently of movement of the soil. Figure 13.25a shows
a schematic diagram of such an arrangement. The bottom of the shafts should be placed below
the active zone of the expansive soil. For the design of the shafts, the uplifting force, U, may be
estimated (see Figure 13.25b) from the equation
U 5 pDsZssw
r tan fps
r
(13.14)
where
Ds 5 diameter of the shaft
Z 5 depth of the active zone
fps
r 5 effective angle of plinth–soil friction
ssw
r 5 pressure for zero swell (see Figures 13.14 and 13.15;
ssw
r 5 sor 1 ssr 1 s1r )
In most cases, the value of fps
r varies between 10 and 20°. An average value of the zero
horizontal swell pressure must be determined in the laboratory. In the absence of laboratory
results, ssw
r tan fps
r may be considered equal to the undrained shear strength of clay, cu , in
the active zone.
714 Chapter 13: Foundations on Difficult Soils
The belled portion of the drilled shaft will act as an anchor to resist the uplifting
force. Ignoring the weight of the drilled shaft, we have
Qnet 5 U 2 D
(13.15)
cuNc p
¢ ≤ (D2b 2 D2s )
FS 4
(13.16)
where
Qnet 5 net uplift load
D 5 dead load
Now,
Qnet <
where
cu 5 undrained cohesion of the clay in which the bell is located
Nc 5 bearing capacity factor
FS 5 factor of safety
Db 5 diameter of the bell of the drilled shaft
Combining Eqs. (13.15) and (13.16) gives
U2D5
cuNc p
¢ ≤ (D2b 2 D2s )
FS 4
(13.17)
Conservatively, from Tables 3.3 and 3.4,
Nc < Nc(strip)Fcs 5 Nc(strip) ¢1 1
NqB
NcL
≤ < 5.14¢1 1
1
≤ 5 6.14
5.14
A drilled-shaft design is examined in Example 13.4.
Example 13.4
Figure 13.26 shows a drilled shaft with a bell. The depth of the active zone is 5 m. The
zero swell pressure of the swelling clay (ssw
r ) is 450 kN> m2. For the drilled shaft, the
dead load (D) is 600 kN and the live load is 300 kN. Assume fps
r 5 12°.
a. Determine the diameter of the bell, Db.
b. Check the bearing capacity of the drilled shaft assuming zero uplift force.
Solution
Part a: Determining the Bell Diameter, Db
The uplift force, Eq. (13.14), is
U 5 pDsZssw
r tan fps
r
13.12 Construction on Expansive Soils
715
Dead load live load 900 kN
Active
zone
5m
800 mm
2m
cu 450 kN/m2
Db
Figure 13.26 Drilled shaft in a swelling clay
Given: Z 5 5 m and ssw
r 5 450 kN>m2. Then
U 5 p(0.8) (5) (450)tan 12° 5 1202 kN
Assume the dead load and live load to be zero, and FS in Eq. (13.17) to be 1.25. So,
from Eq. (13.17),
U5
1202 5
cuNc p
¢ ≤ (D2b 2 D2s )
FS 4
(450) (6.14) p
¢ ≤ (D2b 2 0.82 );
1.25
4
Db 5 1.15 m
The factor of safety against uplift with the dead load also should be checked. A factor of
safety of at least 2 is desirable. So, from Eq. (13.17)
cuNc ¢
FS 5
5
p
≤ (D2b 2 D2s )
4
U2D
p
(450) (6.14) ¢ ≤ (1.152 2 0.82 )
4
1202 2 600
5 2.46 + 2 — OK
Part b: Check for Bearing Capacity
Assume that U 5 0. Then
Dead load 1 live load 5 600 1 300 5 900 kN
Downward load per unit area 5
900
p
¢ ≤ (D2b )
4
5
900
p
¢ ≤ (1.152 )
4
5 866.5 kN>m2
716 Chapter 13: Foundations on Difficult Soils
Net bearing capacity of the soil under the bell 5 qu(net) 5 cuNc
5 (450) (6.14) 5 2763 kN>m2
Hence, the factor of safety against bearing capacity failure is
FS 5
2763
5 3.19 + 3 — OK
866.5
■
Sanitary Landfills
13.13
General Nature of Sanitary Landfills
Sanitary landfills provide a way to dispose of refuse on land without endangering
public health. Sanitary landfills are used in almost all countries, to varying degrees of
success. The refuse disposed of in sanitary landfills may contain organic, wood, paper,
and fibrous wastes, or demolition wastes such as bricks and stones. The refuse is
dumped and compacted at frequent intervals and is then covered with a layer of soil, as
shown in Figure 13.27. In the compacted state, the average unit weight of the
refuse may vary between 5 and 10 kN>m3. A typical city in the United States, with a
population of 1 million, generates about 3.8 3 106 m3 of compacted landfill material
per year.
As property values continue to increase in densely populated areas, constructing
structures over sanitary landfills becomes more and more tempting. In some instances,
a visual site inspection may not be enough to detect an old sanitary landfill.
However, construction of foundations over sanitary landfills is generally problematic
because of poisonous gases (e.g., methane), excessive settlement, and low inherent
bearing capacity.
Soil cover
Excavation
for soil
cover
Landfill
Original
ground
surface
Figure 13.27 Schematic
diagram of a sanitary landfill in progress
13.14 Settlement of Sanitary Landfills
13.14
717
Settlement of Sanitary Landfills
Sanitary landfills undergo large continuous settlements over a long time. Yen and
Scanlon (1975) documented the settlement of several landfill sites in California. After
completion of the landfill, the settlement rate (Figure 13.28) may be expressed as
m5
DHf (m)
Dt (month)
(13.18)
where
m 5 settlement rate
Hf 5 maximum height of the sanitary landfill
On the basis of several field observations, Yen and Scanlon determined the following empirical correlations for the settlement rate:
3for fill heights ranging from 12 to 24 m4
m 5 a 2 b log t1
3for fill heights ranging from 24 to 30 m4
m 5 c 2 d log t1
3for fill heights larger than 30 m4
m 5 e 2 f log t1
In these equations,
(13.19)
(13.20)
(13.21)
m is in m>mo(ft>mo).
t1 is the median fill age, in months
In SI units, the values of a, b, c, d, e, and f given in Eqs. (13.19) through (13.21) are
Item
SI
a
b
c
d
e
f
0.0268
0.0116
0.038
0.0155
0.0435
0.0183
Height of
sanitary
landfill
Hf
Hf
Hf
2
t
t1
tc
2
tc
t t tc
tc t tc
Time, t
Figure 13.28 Settlement of sanitary
landfills
718 Chapter 13: Foundations on Difficult Soils
The median fill age may be defined from Figure 13.28 as
t1 5 t 2
tc
2
(13.22)
where
t 5 time from the beginning of the landfill
tc 5 time for completion of the landfill
Equations (13.19), (13.20), and (13.21) are based on field data from landfills for which
tc varied from 70 to 82 months. To get an idea of the approximate length of time required for
a sanitary landfill to undergo complete settlement, consider Eq. (13.19). For a fill 12 m high
and for tc 5 72 months,
m 5 0.0268 2 0.0116 log t1
so
log t1 5
0.0268 2 m
0.0116
If m 5 0 (zero settlement rate), log t1 5 2.31, or t1 < 200 months. Thus, settlement will
continue for t1 2 tc>2 5 200 2 36 5 164 months (<14 years) after completion of the
fill—a fairly long time. This calculation emphasizes the need to pay close attention to the
settlement of foundations constructed on sanitary landfills.
A comparison of Eqs. (13.19) through (13.21) for rates of settlement shows that
the value of m increases with the height of the fill. However, for fill heights greater than
about 30 m, the rate of settlement should not be much different from that obtained from
Eq. (13.21). The reason is that the decomposition of organic matter close to the surface
is mainly the result of an anaerobic environment. For deeper fills, the decomposition is
slower. Hence, for fill heights greater than about 30 m, the rate of settlement does not
exceed that for fills that are about 30 m in height.
Sowers (1973) also proposed a formula for calculating the settlement of a sanitary
landfill, namely,
DHf 5
aHf
11e
log ¢
ts
≤
tr
(13.23)
where
Hf 5 height of the fill
e 5 void ratio
a 5 a coefficient for settlement
tr, ts 5 times (see Figure 13.28)
DHf 5 settlement between times tr and ts
The coefficients a fall between
a 5 0.09e
(for conditions favorable to decomposition)
(13.24)
Problems
719
and
a 5 0.03e
(for conditions unfavorable to decomposition)
(13.25)
Equation (13.23) is similar to the equation for secondary consolidation settlement.
Problems
13.1 For leossial soil, given Gs 5 2.69. Plot a graph of gd (kN>m3 ) versus liquid limit
to identify the zone in which the soil is likely to collapse on saturation. If a soil has
a liquid limit of 33, Gs 5 2.69, and gd 5 13.5 kN>m3, is collapse likely to occur?
13.2 A collapsible soil layer in the field has a thickness of 3 m. The average effective
overburden pressure on the soil layer is 62 kN>m2. An undisturbed specimen of this
soil was subjected to a double oedometer test. The preconsolidation pressure of the
specimen as determined from the soaked specimen was 84 kN>m2 . Is the soil in the
field normally consolidated or preconsolidated?
13.3 An expansive soil has an active-zone thickness of 8 m. The natural moisture content of the soil is 20% and its liquid limit is 50. Calculate the free surface swell of
the expansive soil upon saturation.
13.4 An expansive soil profile has an active-zone thickness of 5.2 m. A shallow foundation is to be constructed at a depth of 1.2 m below the ground surface. Based on
the swelling pressure test, the following are given.
Depth from ground
surface (m)
1.2
2.2
3.2
4.2
5.2
Swell under overburden and
estimated foundation surcharge
pressure, sw(1) (%)
3.0
2.0
1.2
0.55
0.0
Estimate the total possible swell under the foundation.
13.5 Refer to Problem 13.4. If the allowable total swell is 15 mm., what would be the
necessary undercut?
13.6 Repeat Problem 13.4 with the following: active zone thickness 5 6 m, depth
of shallow foundation 5 1.5 m.
Depth from ground
surface (m)
1.5
2.0
3.0
4.0
5.0
6.0
Swell under overburden and
estimated foundation surcharge
pressure, sw (1) (%)
5.5
3.1
1.5
0.75
0.4
0.0
720 Chapter 13: Foundations on Difficult Soils
13.7 Refer to Problem 13.6. If the allowable total swell is 30 mm, what would be the
necessary undercut?
13.8 Refer to Figure 13.25b. For the drilled shaft with bell, given:
Thickness of active zone, Z 5 9.15 m
Dead load 5 1334 kN
Live load 5 267 kN
Diameter of the shaft, Ds 5 1.07 m
Zero swell pressure for the clay in the active zone 5 574.6 kN>m2
Average angle of plinth-soil friction, fps
r 5 15°
Average undrained cohesion of the clay around the bell 5 144.6 kN>m2
Determine the diameter of the bell, Db. A factor of safety of 3 against uplift is required with the assumption that dead load plus live load is equal to zero.
13.9 Refer to Problem 13.8. If an additional requirement is that the factor of safety
against uplift is at least 3 with the dead load on (live load 5 0), what should be the
diameter of the bell?
References
ABDULJAUWAD, S. N., and AL-SULAIMANI, G. J. (1993). “Determination of Swell Potential of Al-Qatif
Clay,” Geotechnical Testing Journal, American Society for Testing and Materials,Vol. 16,
No. 4, pp. 469 – 484.
ALTMEYER, W. T. (1955). “Discussion of Engineering Properties of Expansive Clays,” Journal of the
Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 81,
No. SM2, pp. 17–19.
BENITES, L. A. (1968). “Geotechnical Properties of the Soils Affected by Piping near the Benson
Area, Cochise County, Arizona,” M. S. Thesis, University of Arizona, Tucson.
CHEN, F. H. (1988). Foundations on Expansive Soils, Elsevier, Amsterdam.
CLEMENCE, S. P., and FINBARR, A. O. (1981). “Design Considerations for Collapsible Soils,” Journal
of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 107,
No. GT3, pp. 305 – 317.
CLEVENGER, W. (1958). “Experience with Loess as Foundation Material,” Transactions, American
Society of Civil Engineers, Vol. 123, pp. 151–170.
DAKSHANAMANTHY, V., and RAMAN, V. (1973). “A Simple Method of Identifying an Expansive Soil,”
Soils and Foundations, Vol. 13, No. 1, pp. 97–104.
DENISOV, N. Y. (1951). The Engineering Properties of Loess and Loess Loams, Gosstroiizdat,
Moscow.
FEDA, J. (1964). “Colloidal Activity, Shrinking and Swelling of Some Clays,” Proceedings, Soil Mechanics Seminar, Loda, Illinois, pp. 531–546.
GIBBS, H. J. (1961). Properties Which Divide Loose and Dense Uncemented Soils, Earth Laboratory
Report EM-658, Bureau of Reclamation, U.S. Department of the Interior, Washington, DC.
GROMKO, G. J. (1974). “Review of Expansive Soils,” Journal of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 100, No. GT6, pp. 667– 687.
HANDY, R. L. (1973). “Collapsible Loess in Iowa,” Proceedings, Soil Science Society of America,
Vol. 37, pp. 281–284.
HOLTZ, W. G. (1959). “Expansive Clays—Properties and Problems,” Journal of the Colorado School
of Mines, Vol. 54, No. 4, pp. 89 –125.
HOLTZ, W. G., and HILF, J. W. (1961). “Settlement of Soil Foundations Due to Saturation,” Proceedings, Fifth International Conference on Soil Mechanics and Foundation Engineering, Paris,
Vol. 1, 1961, pp. 673– 679.
References
721
HOUSTON, W. N., and HOUSTON, S. L. (1989). “State-of-the-Practice Mitigation Measures for Collapsible Soil Sites,” Proceedings, Foundation Engineering: Current Principles and Practices,
American Society of Civil Engineers, Vol. 1, pp. 161–175.
JENNINGS, J. E., and KNIGHT, K. (1975). “A Guide to Construction on or with Materials Exhibiting Additional Settlements Due to ‘Collapse’ of Grain Structure,” Proceedings, Sixth Regional Conference for Africa on Soil Mechanics and Foundation Engineering, Johannesburg, pp. 99–105.
LUTENEGGER, A. J. (1986). “Dynamic Compaction in Friable Loess,” Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 112, No. GT6, pp. 663–667.
LUTENEGGER, A. J., and SABER, R. T. (1988). “Determination of Collapse Potential of Soils,” Geotechnical Testing Journal, American Society for Testing and Materials, Vol. 11. No. 3, pp. 173–178.
MCKEEN, R. G. (1992). “A Model for Predicting Expansive Soil Behavior,” Proceedings, Seventh International Conference on Expansive Soils, Dallas, Vol. 1, pp. 1– 6.
NAYAK, N. V., and CHRISTENSEN, R. W. (1974). “Swell Characteristics of Compacted Expansive
Soils,” Clay and Clay Minerals, Vol. 19, pp. 251–261.
O’NEILL, M. W., and POORMOAYED, N. (1980). “Methodology for Foundations on Expansive Clays,”
Journal of the Geotechnical Engineering Division, American Society of Civil Engineers,
Vol. 106, No. GT12, pp. 1345–1367.
PECK, R. B., HANSON, W. E., and THORNBURN, T. B. (1974). Foundation Engineering, Wiley, New York.
PRIKLONSKII, V. A. (1952). Gruntovedenic, Vtoraya Chast’, Gosgeolzdat, Moscow.
RAMAN, V. (1967). “Identification of Expansive Soils from the Plasticity Index and the Shrinkage Index Data,” The Indian Engineer, Vol. 11, No. 1, pp. 17–22.
SEED, H. B., WOODWARD, R. J., JR., and LUNDGREN, R. (1962). “Prediction of Swelling Potential for
Compacted Clays,” Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 88, No. SM3, pp. 53 –87.
SEMKIN, V. V., ERMOSHIN, V. M., and OKISHEV, N. D. (1986). “Chemical Stabilization of Loess Soils
in Uzbekistan,” Soil Mechanics and Foundation Engineering (trans. from Russian), Vol. 23,
No. 5, pp. 196–199.
SNETHEN, D. R. (1984). “Evaluation of Expedient Methods for Identification and Classification of
Potentially Expansive Soils,” Proceedings, Fifth International Conference on Expansive Soils,
Adelaide, pp. 22–26.
SNETHEN, D. R., JOHNSON, L. D., and PATRICK, D. M. (1977). An Evaluation of Expedient Methodology for Identification of Potentially Expansive Soils, Report No. FHWA-RD-77-94, U.S.
Army Engineers Waterways Experiment Station, Vicksburg, MS.
SOWERS, G. F. (1973). “Settlement of Waste Disposal Fills,” Proceedings, Eighth International Conference on Soil Mechanics and Foundation Engineering, Moscow, pp. 207–210.
SOWERS, G. B., and SOWERS, G. F. (1970). Introductory Soil Mechanics and Foundations, 3d ed.
Macmillan, New York.
SRIDHARAN, A. (2005). “On Swelling Behaviour of Clays,” Proceedings, International Conference
on Problematic Soils, North Cyprus, Vol. 2, pp. 499–516.
SRIDHARAN, A., RAO, A. S., and SIVAPULLAIAH, P. V. (1986), “Swelling Pressure of Clays,” Geotechnical
Testing Journal, American Society for Testing and Materials, Vol. 9, No. 1, pp. 24–33.
Uniform Building Code (1968). UBC Standard No. 29-2.
VAN DER MERWE, D. H. (1964), “The Prediction of Heave from the Plasticity Index and Percentage
Clay Fraction of Soils,” Civil Engineer in South Africa, Vol. 6, No. 6, pp. 103–106.
VIJAYVERGIYA, V. N., and GHAZZALY, O. I. (1973). “Prediction of Swelling Potential of Natural
Clays,” Proceedings, Third International Research and Engineering Conference on Expansive
Clays, pp. 227–234.
WESTON, D. J. (1980). “Expansive Roadbed Treatment for Southern Africa,” Proceedings, Fourth International Conference on Expansive Soils, Vol. 1, pp. 339–360.
YEN, B. C., and SCANLON, B. (1975). “Sanitary Landfill Settlement Rates,” Journal of the Geotechnical
Engineering Division, American Society of Civil Engineers, Vol. 101, No. GT5, pp. 475–487.
14
14.1
Soil Improvement and Ground Modification
Introduction
The soil at a construction site may not always be totally suitable for supporting
structures such as buildings, bridges, highways, and dams. For example, in granular soil
deposits, the in situ soil may be very loose and indicate a large elastic settlement. In such
a case, the soil needs to be densified to increase its unit weight and thus its shear
strength.
Sometimes the top layers of soil are undesirable and must be removed and
replaced with better soil on which the structural foundation can be built. The soil used
as fill should be well compacted to sustain the desired structural load. Compacted fills
may also be required in low-lying areas to raise the ground elevation for construction of
the foundation.
Soft saturated clay layers are often encountered at shallow depths below foundations. Depending on the structural load and the depth of the layers, unusually large consolidation settlement may occur. Special soil-improvement techniques are required to
minimize settlement.
In Chapter 13, we mentioned that the properties of expansive soils could be altered
substantially by adding stabilizing agents such as lime. Improving in situ soils by using
additives is usually referred to as stabilization.
Various techniques are used to
1. Reduce the settlement of structures
2. Improve the shear strength of soil and thus increase the bearing capacity of shallow
foundations
3. Increase the factor of safety against possible slope failure of embankments
and earth dams
4. Reduce the shrinkage and swelling of soils
This chapter discusses some of the general principles of soil improvement, such as compaction, vibroflotation, precompression, sand drains, wick drains, stabilization by admixtures, jet grouting, and deep mixing, as well as the use of stone columns and sand
compaction piles in weak clay to construct foundations.
722
14.2 General Principles of Compaction
14.2
723
General Principles of Compaction
If a small amount of water is added to a soil that is then compacted, the soil will have
a certain unit weight. If the moisture content of the same soil is gradually increased and
the energy of compaction is the same, the dry unit weight of the soil will gradually
increase. The reason is that water acts as a lubricant between the soil particles, and
under compaction it helps rearrange the solid particles into a denser state. The increase
in dry unit weight with increase of moisture content for a soil will reach a limiting
value beyond which the further addition of water to the soil will result in a reduction
in dry unit weight. The moisture content at which the maximum dry unit weight is
obtained is referred to as the optimum moisture content.
The standard laboratory tests used to evaluate maximum dry unit weights and optimum moisture contents for various soils are
• The Standard Proctor test (ASTM designation D-698)
• The Modified Proctor test (ASTM designation D-1557)
The soil is compacted in a mold in several layers by a hammer. The moisture content of the soil,
w, is changed, and the dry unit weight, gd , of compaction for each test is determined. The maximum dry unit weight of compaction and the corresponding optimum moisture content are
determined by plotting a graph of gd against w (%). The standard specifications for the two
types of Proctor test are given in Tables 14.1 and 14.2.
Table 14.1 Specifications for Standard Proctor Test (Based on ASTM Designation 698)
Item
Method A
Method B
Method C
Diameter of mold
Volume of mold
Mass of hammer
Height of hammer drop
Number of hammer
blows per layer of soil
Number of layers of
compaction
Energy of compaction
Soil to be used
101.6 mm
944 cm3
2.5 kg
304.8 mm
25
101.6 mm
944 cm3
2.5 kg
304.8 mm
25
152.4 mm
2124 cm3
2.5 kg
304.8 mm
56
3
3
3
600 kN # m>m3
Portion passing No. 4
4.57-mm sieve.
May be used if 20% or
less by weight of material
is retained on No. 4 sieve.
600 kN # m>m3
Portion passing
9.5-mm sieve.
May be used if soil
retained on No. 4 sieve
is more than 20% and
20% or less by weight
is retained on 9.5-mm
(38 -in.) sieve.
600 kN # m>m3
Portion passing
19.0-mm sieve.
May be used if more
than 20% by weight of
material is retained on
9.5 mm sieve
and less than 30% by
weight is retained on
19.00-mm sieve.
724 Chapter 14: Soil Improvement and Ground Modification
Table 14.2 Specifications for Modified Proctor Test (Based on ASTM Designation 1557)
Item
Method A
Method B
Method C
Diameter of mold
Volume of mold
Mass of
hammer
Height of
hammer drop
Number of hammer
blows per layer
of soil
Number of layers
of compaction
Energy of
compaction
Soil to be used
101.6 mm
944 cm3
4.54 kg
101.6 mm
944 cm3
4.54 kg
152.4 mm
2124 cm3
4.54 kg
457.2 mm
457.2 mm
457.2 mm
25
25
56
5
5
5
2700 kN # m>m3
2700 kN # m>m3
2700 kN # m>m3
Portion passing
No. 4 (4.57-mm)
sieve. May be used
if 20% or less by
weight of material
is retained on
No. 4 sieve.
Portion passing
9.5-mm sieve.
May be used if soil
retained on No. 4
sieve is more than
20% and 20% or
less by weight is retained on 9.5-mm
sieve.
Portion passing 19.0-mm
(34 -in.) sieve.
May be used if more
than 20% by weight
of material is retained
on 9.5-mm
sieve and less than
30% by weight is
retained on 19-mm
sieve.
Figure 14.1 shows a plot of gd against w (%) for a clayey silt obtained from standard and modified Proctor tests (method A). The following conclusions may be drawn:
1. The maximum dry unit weight and the optimum moisture content depend on the
degree of compaction.
2. The higher the energy of compaction, the higher is the maximum dry unit weight.
3. The higher the energy of compaction, the lower is the optimum moisture content.
4. No portion of the compaction curve can lie to the right of the zero-air-void line. The
zero-air-void dry unit weight, gzav , at a given moisture content is the theoretical
maximum value of gd , which means that all the void spaces of the compacted soil
are filled with water, or
gzav 5
where
gw 5 unit weight of water
Gs 5 specific gravity of the soil solids
w 5 moisture content of the soil
gw
1
1w
Gs
(14.1)
14.2 General Principles of Compaction
725
24
Dry unit weight, ␥d (kN/m3)
22
Zero-air-void
curve
(Gs 2.7)
20
18
16
14
Standard
Proctor test
12
Modified
Proctor test
10
0
5
10
15
20
Moisture content, w (%)
25
Figure 14.1 Standard and modified
Proctor compaction curves for a clayey
silt (method A)
5. The maximum dry unit weight of compaction and the corresponding optimum moisture content will vary from soil to soil.
Using the results of laboratory compaction (gd versus w), specifications may be
written for the compaction of a given soil in the field. In most cases, the contractor is
required to achieve a relative compaction of 90% or more on the basis of a specific laboratory test (either the standard or the modified Proctor compaction test). The relative compaction is defined as
RC 5
gd(field)
(14.2)
gd(max)
Chapter 1 introduced the concept of relative density (for the compaction of granular
soils), defined as
Dr 5 B
gd 2 gd(min)
gd(max) 2 gd(min)
R
gd(max)
gd
726 Chapter 14: Soil Improvement and Ground Modification
where
gd 5 dry unit weight of compaction in the field
gd(max) 5 maximum dry unit weight of compaction as determined in the laboratory
gd(min) 5 minimum dry unit weight of compaction as determined in the laboratory
For granular soils in the field, the degree of compaction obtained is often measured in terms
of relative density. Comparing the expressions for relative density and relative compaction
reveals that
RC 5
where A 5
gd(min)
gd(max)
A
1 2 Dr (1 2 A)
(14.3)
.
Omar, et al. (2003) recently presented the results of modified Proctor compaction tests
on 311 soil samples. Of these samples, 45 were gravelly soil (GP, GP-GM, GW, GW-GM, and
GM), 264 were sandy soil (SP, SP-SM, SW-SM, SW, SC-SM, SC, and SM), and two were
clay with low plasticity (CL). All compaction tests were conducted using ASTM 1557 method
C to avoid over-size correction. Based on the tests, the following correlations were developed.
rd(max ) (kg>m3 ) 5 34,804,574GS 2 195.55(LL) 2 1 156,971(R[ 4) 0.5
2 9,527,8304 0.5
24
2
(14.4)
25
ln(wopt ) 5 1.195 3 10 (LL) 2 1.964Gs 2 6.617 3 10 (R[ 4)
1 7.651
(14.5)
where
␳d(max) ⫽ maximum dry density
wopt ⫽ optimum moisture content
Gs ⫽ specific gravity of soil solids
LL ⫽ liquid limit, in percent
R # 4 ⫽ percent retained on No. 4 sieve
It needs to be pointed out that Eqs. (14.4 and 14.5) contain the term for liquid limit.
This is because the soils that were considered included silty and clayey sands.
Osman et al. (2008) analyzed a number of laboratory compaction-test results on finegrained (cohesive) soil. Based on this study, the following correlations were developed:
wopt 5 (1.99 2 0.165 ln E) (PI)
(14.6)
gd(max) 5 L 2 Mwopt
(14.7)
and
14.3 Field Compaction
727
where
L ⫽ 14.34 ⫹ 1.195 ln E
M ⫽ ⫺0.19 ⫹ 0.073 ln E
(14.8)
(14.9)
wopt ⫽ optimum moisture content (%)
PI ⫽ plasticity index (%)
␥d(max) ⫽ maximum dry unit weight (kN/m3)
E ⫽ compaction energy (kN-m/m3)
14.3
Field Compaction
Ordinary compaction in the field is done by rollers. Of the several types of roller used, the
most common are
1.
2.
3.
4.
Smooth-wheel rollers (or smooth drum rollers)
Pneumatic rubber-tired rollers
Sheepsfoot rollers
Vibratory rollers
Figure 14.2 shows a smooth-wheel roller that can also create vertical vibration
during compaction. Smooth-wheel rollers are suitable for proof-rolling subgrades
and for finishing the construction of fills with sandy or clayey soils. They provide
100% coverage under the wheels, and the contact pressure can be as high as
Figure 14.2 Vibratory smooth-wheel rollers (Courtesy of Tampo Manufacturing Co., Inc.,
San Antonio, Texas)
728 Chapter 14: Soil Improvement and Ground Modification
Figure 14.3 Pneumatic rubber-tired roller (Courtesy of Tampo Manufacturing Co., Inc.,
San Antonio, Texas)
300 to 400 kN>m2. However, they do not produce a uniform unit weight of compaction
when used on thick layers.
Pneumatic rubber-tired rollers (Figure 14.3) are better in many respects than
smooth-wheel rollers. Pneumatic rollers, which may weigh as much as 2000 kN, consist of a heavily loaded wagon with several rows of tires. The tires are closely spaced—
four to six in a row. The contact pressure under the tires may range up to
600 to 700 kN>m2, and they give about 70 to 80% coverage. Pneumatic rollers, which
can be used for sandy and clayey soil compaction, produce a combination of pressure
and kneading action.
Sheepsfoot rollers (Figure 14.4) consist basically of drums with large numbers
of projections. The area of each of the projections may be 25 to 90 cm2. These rollers
are most effective in compacting cohesive soils. The contact pressure under the projections may range from 1500 to 7500 kN>m2. During compaction in the field, the initial passes compact the lower portion of a lift. Later, the middle and top of the lift are
compacted.
Vibratory rollers are efficient in compacting granular soils. Vibrators can be attached to
smooth-wheel, pneumatic rubber-tired or sheepsfoot rollers to send vibrations into the soil
being compacted. Figures 14.2 and 14.4 show vibratory smooth-wheel rollers and a vibratory
sheepsfoot roller, respectively.
In general, compaction in the field depends on several factors, such as the type of
compactor, type of soil, moisture content, lift thickness, towing speed of the compactor,
and number of passes the roller makes.
Figure 14.5 shows the variation of the unit weight of compaction with depth
for a poorly graded dune sand compacted by a vibratory drum roller. Vibration was
14.3 Field Compaction
729
Figure 14.4 Vibratory sheepsfoot roller (Courtesy of Tampo Manufacturing Co., Inc.,
San Antonio, Texas)
16
Dry unit weight (kN/m3)
16.5
17
0
Depth (m)
0.5
1.0
2
5
15
1.5
45 Number
of roller
passes
2.0
Figure 14.5 Vibratory compaction of a sand:
Variation of dry unit weight with depth and
number of roller passes; lift thickness 5 2.44 m
(After D’Appolonia et al., 1969) (After
D’Appolonia, D. J., Whitman, R. V. and
D’Appolonia, E. (1969). “Sand Compaction with
Vibratory Rollers,” Journal of the Soil Mechanics
and Foundations Division, American Society of
Civil Engineers, Vol. 95, N. SM1, pp. 263–284.
With permission from ASCE.)
730 Chapter 14: Soil Improvement and Ground Modification
produced by mounting an eccentric weight on a single rotating shaft within the
drum cylinder. The weight of the roller used for this compaction was 55.7 kN, and
the drum diameter was 1.19 m. The lifts were kept at 2.44 m. Note that, at any depth,
the dry unit weight of compaction increases with the number of passes the roller makes.
However, the rate of increase in unit weight gradually decreases after about 15 passes.
Note also the variation of dry unit weight with depth by the number of roller
passes. The dry unit weight and hence the relative density, Dr , reach maximum values
at a depth of about 0.5 m and then gradually decrease as the depth increases. The reason is the lack of confining pressure toward the surface. Once the relation between
depth and relative density (or dry unit weight) for a soil for a given number of passes
is determined, for satisfactory compaction based on a given specification, the approximate thickness of each lift can be easily estimated.
Hand-held vibrating plates can be used for effective compaction of granular soils
over a limited area. Vibrating plates are also gang-mounted on machines. These can be
used in less restricted areas.
14.4
Compaction Control for Clay
Hydraulic Barriers
Compacted clays are commonly used as hydraulic barriers in cores of earth dams, liners
and covers of landfills, and liners of surface impoundments. Since the primary purpose
of a barrier is to minimize flow, the hydraulic conductivity, k, is the controlling factor.
In many cases, it is desired that the hydraulic conductivity be less than 1027cm>s. This
can be achieved by controlling the minimum degree of saturation during compaction, a
relation that can be explained by referring to the compaction characteristics of three soils
described in Table 14.3 (Othman and Luettich, 1994).
Figures 14.6, 14.7, and 14.8 show the standard and modified Proctor test results
and the hydraulic conductivities of compacted specimens. Note that the solid symbols
represent specimens with hydraulic conductivities of 1027cm>s or less. As can be seen
from these figures, the data points plot generally parallel to the line of full saturation.
Figure 14.9 shows the effect of the degree of saturation during compaction on the
hydraulic conductivity of the three soils. It is evident from the figure that, if it is desired
that the maximum hydraulic conductivity be 1027cm>s, then all soils should be compacted at a minimum degree of saturation of 88%.
Table 14.3 Characteristics of Soils Reported in Figures 14.6, 14.7, and 14.8
Soil
Classification
Liquid
limit
Wisconsin A
Wisconsin B
Wisconsin C
CL
CL
CH
34
42
84
Plasticity
index
Percent finer
than No. 200
sieve (0.075 mm)
16
19
60
85
99
71
14.4 Compaction Control for Clay Hydraulic Barriers
731
Hydraulic conductivity (cm/s)
10–5
Standard Proctor
Modified Proctor
10–6
10–7
10–8
10–9
10
12
14
16
18
Moisture content (%)
(a)
20
22
24
Dry unit weight (kN/m3)
19
18
Sa
tur
ati
on
17
10
0%
90
%
16
80
%
Solid symbols represent specimens with hydraulic
conductivity equal to or less than 1 10–7 cm/s
15
10
12
14
16
18
Moisture content (%)
(b)
20
22
24
Figure 14.6 Standard and Modified Proctor test results and hydraulic conductivity of Wisconsin
A soil (After Othman and Luettich, 1994) (From Othman, M. A., and S. M. Luettich. Compaction
Control Criteria for Clay Hydraulic Barriers. In Transportation Research Record 1462,
Transportation Research Board, National Research Council, Washington, D.C., 1994,
Figures 4 and 5, p. 32, and Figures 6 and 7, p. 33. Reproduced with permission of the
Transportation Research Board.)
In field compaction at a given site, soils of various composition may be encountered.
Small changes in the content of fines will change the magnitude of hydraulic conductivity.
Hence, considering the various soils likely to be encountered at a given site the procedure just
described aids in developing a minimum-degree-of-saturation criterion for compaction to construct hydraulic barriers.
732 Chapter 14: Soil Improvement and Ground Modification
10–5
Hydraulic conductivity (cm/s)
Solid symbols represent specimens with hydraulic
conductivity equal to or less than 1 10–7 cm/s
10–6
10–7
10–8
10–9
8
12
16
Moisture content (%)
(a)
20
24
19
Figure 14.7 Standard and Modified
Proctor test results and hydraulic conductivity of Wisconsin B soil (After
Othman and Luettich, 1994) (From
Othman, M. A., and S. M. Luettich.
Compaction Control Criteria for Clay
Hydraulic Barriers. In Transportation
Research Record 1462, Transportation
Research Board, National Research
Council, Washington, D.C., 1994,
Figures 4 and 5, p. 32, and Figures 6
and 7, p. 33. Reproduced with permission of the Transportation Research
Board.)
14.5
Dry unit weight (kN/m3)
Standard Proctor
Modified Proctor
18
Sa
17
tur
ati
o
n
10
0%
16
90
%
80
%
15
8
12
16
Moisture content (%)
(b)
20
24
Vibroflotation
Vibroflotation is a technique developed in Germany in the 1930s for in situ densification of
thick layers of loose granular soil deposits. Vibroflotation was first used in the United States
about 10 years later. The process involves the use of a vibroflot (called the vibrating unit),
as shown in Figure 14.10. The device is about 2 m in length. This vibrating unit has an eccentric weight inside it and can develop a centrifugal force. The weight enables the unit to vibrate
horizontally. Openings at the bottom and top of the unit are for water jets. The vibrating unit
is attached to a follow-up pipe. The figure shows the vibroflotation equipment necessary for
compaction in the field.
14.5 Vibroflotation
733
Hydraulic conductivity (cm/s)
10–5
Solid symbols represent
specimens with hydraulic
conductivity equal to or
less than 1 10–7 cm/s
10–6
10–7
10–8
10–9
15
20
25
Moisture content (%)
(a)
30
35
16.51
Standard Proctor
Modified Proctor
Figure 14.8 Standard and Modified
Proctor test results and hydraulic conductivity of Wisconsin C soil (After
Othman and Luettich, 1994) (From
Othman, M. A., and S. M. Luettich.
Compaction Control Criteria for Clay
Hydraulic Barriers. In Transportation
Research Record 1462, Transportation
Research Board, National Research
Council, Washington, D.C., 1994,
Figures 4 and 5, p. 32, and Figures 6 and
7, p. 33. Reproduced with permission of
the Transportation Research Board.)
Dry unit weight (kN/m3)
16.0
Sa
tu
15.0
ra
ti
on
%
%
15
20
25
Moisture content (%)
(b)
10
0%
90
80
14.0
30
35
The entire compaction process can be divided into four steps (see Figure 14.11):
Step 1. The jet at the bottom of the vibroflot is turned on, and the vibroflot is lowered into the ground.
Step 2. The water jet creates a quick condition in the soil, which allows the vibrating unit to sink.
Step 3. Granular material is poured into the top of the hole. The water from the
lower jet is transferred to the jet at the top of the vibrating unit. This water
carries the granular material down the hole.
Step 4. The vibrating unit is gradually raised in about 0.3-m lifts and is held vibrating for about 30 seconds at a time. This process compacts the soil to the
desired unit weight.
734 Chapter 14: Soil Improvement and Ground Modification
Hydraulic conductivity (cm/s)
10–5
10–6
10–7
Soil A, Standard Proctor
Soil A, Modified Proctor
Soil B, Standard Proctor
Soil B, Modified Proctor
Soil C, Standard Proctor
Soil C, Modified Proctor
10–8
10–9
40
50
60
70
80
Degree of saturation (%)
90
100
Figure 14.9 Effect of degree of saturation on hydraulic conductivity of Wisconsin A, B, and C soils
(After Othman and Luettich, 1994) (From Othman, M. A., and S. M. Luettich. Compaction Control
Criteria for Clay Hydraulic Barriers. In Transportation Research Record 1462, Transportation
Research Board, National Research Council, Washington, D.C., 1994, Figures 4 and 5, p. 32, and
Figures 6 and 7, p. 33. Reproduced with permission of the Transportation Research Board.)
Table 14.4 gives the details of various types of vibroflot unit used in the United States.
The 23 kW electric units have been used since the latter part of the 1940s. The 100-HP units
were introduced in the early 1970s. The zone of compaction around a single probe will vary
according to the type of vibroflot used. The cylindrical zone of compaction will have a
radius of about 2 m for a 23 kW unit and about 3 m for a 75 kW unit. Compaction by
vibroflotation involves various probe spacings, depending on the zone of compaction.
(See Figure 14.12.) Mitchell (1970) and Brown (1977) reported several successful cases of
foundation design that used vibroflotation.
The success of densification of in situ soil depends on several factors, the most important of which are the grain-size distribution of the soil and the nature of the backfill used to fill
the holes during the withdrawal period of the vibroflot. The range of the grain-size distribution
of in situ soil marked Zone 1 in Figure 14.13 is most suitable for compaction by vibroflotation.
Soils that contain excessive amounts of fine sand and silt-size particles are difficult to compact;
for such soils, considerable effort is needed to reach the proper relative density of compaction.
Zone 2 in Figure 14.13 is the approximate lower limit of grain-size distribution for compaction
by vibroflotation. Soil deposits whose grain-size distribution falls into Zone 3 contain appreciable amounts of gravel. For these soils, the rate of probe penetration may be rather slow, so
compaction by vibroflotation might prove to be uneconomical in the long run.
The grain-size distribution of the backfill material is one of the factors that control
the rate of densification. Brown (1977) defined a quantity called suitability number for rating a backfill material. The suitability number is given by the formula
3
1
1
SN 5 1.7
1
1
2
2
Å (D50 )
(D20 )
(D10 ) 2
(14.10)
14.5 Vibroflotation
735
Power
supply
Water
pump
Follow-up
pipe
Vibrating
unit
A
A
Cylinder of compacted
material, added from the
suface to compensate
for the loss of volume
caused by the increase in
density of the compacted
soil
B
Cylinder of compacted
material, produced by a
single vibroflot compaction
B
Figure 14.10 Vibroflotation unit (After Brown, 1977. With permission from ASCE.)
where D50 , D20 , and D10 are the diameters (in mm) through which 50%, 20%, and 10%, respectively, of the material is passing. The smaller the value of SN, more desirable is the
backfill material. Following is a backfill rating system proposed by Brown (1977):
Range of SN
0–10
10–20
20–30
30–50
.50
Rating as backfill
Excellent
Good
Fair
Poor
Unsuitable
736 Chapter 14: Soil Improvement and Ground Modification
Step 1
Step 2
Step 3
Step 4
Figure 14.11 Compaction by the vibroflotation process (After Brown, 1977. With permission
from ASCE.)
Table 14.4 Types of Vibrating Unitsa
75 kW electric and
hydraulic motors
23 kW electric motors
(a) Vibrating tip
Length
Diameter
Weight
Maximum movement when free
Centrifugal force
2.1 m
406 mm
18 kN
12.5 mm
160 kN
1.86 m
381 mm
18 kN
7.6 mm
90 kN
(b) Eccentric
Weight
Offset
Length
Speed
1.16 kN
38 mm
610 mm
1800 rpm
0.76 kN
32 mm
387 mm
1800 rpm
(c) Pump
Operating flow rate
Pressure
0 –1.6 m3>min
690–1035 kN>m2
0–0.6 m3> min
690–1035 kN> m2
(d) Lower follow-up pipe and extensions
Diameter
Weight
a
305 mm
3.65 kN>m
305 mm
3.65 kN>m
After Brown, R. E. (1977), “Vibroflotation Compaction of Cohesionless Soils,” Journal of the
Geotechnical Engineering Division, Vol. 103, No. GT12. With permission from ASCE.
14.5 Vibroflotation
737
Probe
spacing
Zone of influence
for each probe
Figure 14.12 Nature of probe spacing for vibroflotation
Unified Soil Classification System
Sand
Silts and clays
Gravel
100
Percent finer
80
60
Zone 3
Zone 1
Zone 2
40
20
0
100
10
1
0.1
Grain size (mm)
0.01
0.001
Figure 14.13 Effective range of grain-size distribution of soil for vibroflotation
An excellent case study that evaluated the benefits of vibroflotation was presented
by Basore and Boitano (1969). Densification of granular subsoil was necessary for the
construction of a three-story office building at the Treasure Island Naval Station in San
Francisco, California. The top 9 m of soil at the site was loose to medium-dense sand fill
that had to be compacted. Figure 14.14a shows the nature of the layout of the vibroflotation points. Sixteen compaction points were arranged in groups of four, with 1.22 m,
1.52 m, 1.83 m, and 2.44 m spacing. Prior to compaction, standard penetration tests were
conducted at the centers of groups of three compaction points. After the completion of
compaction by vibroflotation, the variation of the standard penetration resistance with
depth was determined at the same points.
738 Chapter 14: Soil Improvement and Ground Modification
S
S
Standard
penetration
test
points
Vibroflotation
point
S
S
(a)
Standard penetration resistance, N60
0
20
40
60
0
Depth (m)
2
4
S 1.22 m—After
Compaction
S 1.22 m—Before
Compaction
S 2.44 m—After
Compaction
S 2.44 m—Before
Compaction
6
8
9
(b)
Figure 14.14 (a) Layout of vibroflotation compaction points; (b) variation of standard penetration resistance (N60) before and after compaction (Basore, C. E. and Boitano, J. D. (1969). “Sand
Densification by Piles and Vibrofloation,” Journal of Soil Mechanics and Foundation Engineering
Division, American Society of Civil Engineers, Vol. 95, No. 6, pp. 1301–1323, Figure 16. With
permission from ASCE.)
Figure 14.14b shows the variation of standard penetration resistance, N60,
with depth before and after compaction for vibroflotation point spacings Sr 5 1.22 m
and 2.44 m. From this figure, the following general conclusions can be drawn:
• For any given Sr, the magnitude of N60 after compaction decreases with an
increase in depth.
• An increase in N60 indicates an increase in the relative density of sand.
14.6 Blasting
739
• The degree of compaction decreases with the increase in Sr. At Sr 5 1.22 m, the
degree of compaction at any depth is the largest. However, at Sr 5 2.44 m, the
vibroflotation had practically no effect in compacting soil.
During the past 30 to 35 years, the vibroflotation technique has been used successfully on large projects to compact granular subsoils, thereby controlling structural
settlement.
14.6
Blasting
Blasting is a technique that has been used successfully in many projects (Mitchell,
1970) for the densification of granular soils. The general soil grain sizes suitable
for compaction by blasting are the same as those for compaction by vibroflotation.
The process involves the detonation of explosive charges such as 60% dynamite at a
certain depth below the ground surface in saturated soil. The lateral spacing of the
charges varies from about 3 to 9 m. Three to five successful detonations are usually
necessary to achieve the desired compaction. Compaction (up to a relative density of
about 80%) up to a depth of about 18 m over a large area can easily be achieved
by using this process. Usually, the explosive charges are placed at a depth of about twothirds of the thickness of the soil layer desired to be compacted. The sphere of
influence of compaction by a 60% dynamite charge can be given as follows
(Mitchell, 1970):
r5
WEX
É C
(14.11)
where r ⫽ sphere of influence
WEX ⫽ weight of explosive ⫺60% dynamite
C ⫽ 0.0122 when WEX is in kg and r is in m
Figure 14.15 shows the test results of soil densification by blasting in an area
measuring 15 m by 9 m (Mitchell, 1970). For these tests, twenty 2.09-kg charges of
Gelamite No. 1 (Hercules Powder Company, Wilmington, Delaware) were used.
14.7
Precompression
When highly compressible, normally consolidated clayey soil layers lie at a limited depth
and large consolidation settlements are expected as the result of the construction of large
buildings, highway embankments, or earth dams, precompression of soil may be used to
minimize postconstruction settlement. The principles of precompression are best explained
740 Chapter 14: Soil Improvement and Ground Modification
1.0
Marker No.
M2
Test No.
3
0.8
Settlement (ft)
M2
0.6
4
M1
4
0.4
M1
3
15 m
0.2
9m
M1
M2
0.0
0
5
10
15
Number of charges
20
25
Figure 14.15 Ground settlement as a function of
number of explosive charges
r , and
by reference to Figure 14.16. Here, the proposed structural load per unit area is Ds(p)
the thickness of the clay layer undergoing consolidation is Hc . The maximum primary consolidation settlement caused by the structural load is then
Sc(p) 5
sor 1 Ds(p)
r
CcHc
log
1 1 eo
sor
(14.12)
The settlement–time relationship under the structural load will be like that shown in
r 1 Ds(f)
r is placed on the ground, the
Figure 14.16b. However, if a surcharge of Ds(p)
primary consolidation settlement will be
Sc(p1f) 5
sor 1 3Ds(p)
r 1 Ds(f)
r 4
CcHc
log
1 1 eo
sor
(14.13)
r 1 Ds(f)
r is also shown in
The settlement–time relationship under a surcharge of Ds(p)
Figure 14.16b. Note that a total settlement of Sc(p) would occur at time t2 , which is much
14.7 Precompression
Surcharge
per unit area
Surcharge
Groundwater table
␴(p) ␴(f )
␴(p)
Sand
Time
t2
Hc
741
t1
Clay
Time
Sc (p)
Sc (p f )
Sand
(a)
Settlement
(b)
Figure 14.16 Principles of precompression
r 1 Ds(f)
r is applied on the
shorter than t1 . So, if a temporary total surcharge of Ds(p)
ground surface for time t2 , the settlement will equal Sc(p) . At that time, if the surcharge is
r is built, no appreremoved and a structure with a permanent load per unit area of Ds(p)
ciable settlement will occur. The procedure just described is called precompression. The
r 1 Ds(f)
r can be applied by means of temporary fills.
total surcharge Ds(p)
Derivation of Equations for Obtaining Ds9(f ) and t2
r 1 Ds(f)
r , the degree of consolidaFigure 14.16b shows that, under a surcharge of Ds(p)
tion at time t2 after the application of the load is
U5
Sc(p)
(14.14)
Sc(p1f)
Substitution of Eqs. (14.12) and (14.13) into Eq. (14.14) yields
log B
U5
log B
r
sor 1 Ds(p)
sor
R
sor 1 Ds(r p) 1 Ds(r f)
sor
log B1 1
5
R
log b1 1
Ds(p)
r
sor
r
Ds(p)
sor
B1 1
R
Ds(f
r )
Ds(p
r )
(14.15)
Rr
r >sor , and
Figure 14.17 gives magnitudes of U for various combinations of Ds(p)
Ds(f)
r >Ds(p)
r . The degree of consolidation referred to in Eq. (14.15) is actually the average
degree of consolidation at time t2 , as shown in Figure 14.17b. However, if the average degree
742 Chapter 14: Soil Improvement and Ground Modification
100
90
␴(p)
=
␴o
8.0
10.0
5.0 6.0
3.0 4.0
2.0
1.4
1.0
0.5
0.3
80
70
U(%)
60
50
40
30
0.1
0
0.2
0.4
0.6
0.8
1.0
␴(f )
␴(p)
1.2
1.4
1.6
1.8
2.0
r > Ds(p)
r for various values of Ds(p)
r >sor —
Figure 14.17 Plot of U against Ds(f)
Eq. (14.15)
of consolidation is used to determine t2 , some construction problems might occur. The reason
is that, after the removal of the surcharge and placement of the structural load, the portion of
clay close to the drainage surface will continue to swell, and the soil close to the midplane
will continue to settle. (See Figure 14.18.) In some cases, net continuous settlement might
result. A conservative approach may solve the problem; that is, assume that U in Eq. (14.15)
is the midplane degree of consolidation (Johnson, 1970a). Now, from Eq. (1.73),
U 5 f(Tv )
Sand
Degree of
consolidation
(decreasing)
100 %
Hc /2
Clay
Hc
Degree of
consolidation
Midplane
Uav
Hc /2
Sand
Depth
Figure 14.18
(1.73)
14.7 Precompression
743
where
Tv 5 time factor 5 Cvt2>H 2
Cv 5 coefficient of consolidation
t2 5 time
H 5 maximum drainage path (5Hc>2 for two-way drainage and Hc for one-way
drainage)
The variation of U (the midplane degree of consolidation) with Tv is given in Figure 14.19.
Procedure for Obtaining Precompression Parameters
Two problems may be encountered by engineers during precompression work in the field:
1. The value of Ds(f)
r is known, but t2 must be obtained. In such a case, obtain sor , Ds(p) ,
and solve for U, using Eq. (14.15) or Figure 14.17. For this value of U, obtain Tv from
Figure 14.19. Then
t2 5
TvH 2
Cv
(14.16)
0
10
20
Degree of consolidation, U (%)
30
40
50
60
70
80
90
100
0.1
0.3
1.0
Tv
2.0
Figure 14.19 Plot of midplane
degree of consolidation against Tv
744 Chapter 14: Soil Improvement and Ground Modification
2. For a specified value of t2 , Ds(f)
r must be obtained. In such a case, calculate Tv .
Then use Figure 14.19 to obtain the midplane degree of consolidation, U. With the
estimated value of U, go to Figure 14.17 to get the required value of Ds(f)
r > Ds(p)
r ,
and then calculate Ds(f)
r .
Several case histories on the successful use of precompression techniques for improving foundation soil are available in the literature (for example, Johnson, 1970a).
Example 14.1
Examine Figure 14.16. During the construction of a highway bridge, the average permanent load on the clay layer is expected to increase by about 115 kN>m2. The average
effective overburden pressure at the middle of the clay layer is 210 kN>m2. Here,
Hc 5 6 m, Cc 5 0.28, eo 5 0.9, and Cv 5 0.36 m2>mo. The clay is normally consolidated. Determine
a. The total primary consolidation settlement of the bridge without precompression
b. The surcharge, Ds(f)
r , needed to eliminate the entire primary consolidation settlement in nine months by precompression.
Solution
Part a
The total primary consolidation settlement may be calculated from Eq. (14.12):
Sc(p) 5
r
sor 1 Ds(p)
CcHc
(0.28) (6)
210 1 115
log B
R 5
log B
R
1 1 eo
sor
1 1 0.9
210
5 0.1677 m 5 167.7 mm
Part b
We have
Tv 5
Cv t2
H2
Cv 5 0.36 m2>mo.
H 5 3 m (two-way drainage)
t2 5 9 mo.
Hence,
Tv 5
(0.36) (9)
32
5 0.36
According to Figure 14.19, for Tv 5 0.36, the value of U is 47%. Now,
r 5 115 kN>m2
Ds(p)
14.8 Sand Drains
745
and
sor 5 210 kN>m2
so
Ds(p)
r
sor
5
115
5 0.548
210
r >sor 5 0.548, Ds(f)
r >Ds(p)
r < 1.8;
According to Figure 14.17, for U 5 47% and Ds(p)
thus,
Ds(f)
r 5 (1.8) (115) 5 207 kN , m2
14.8
■
Sand Drains
The use of sand drains is another way to accelerate the consolidation settlement of soft, normally consolidated clay layers and achieve precompression before the construction of a
desired foundation. Sand drains are constructed by drilling holes through the clay layer(s)
in the field at regular intervals. The holes are then backfilled with sand. This can be achieved
by several means, such as (a) rotary drilling and then backfilling with sand; (b) drilling by
continuous-flight auger with a hollow stem and backfilling with sand (through the hollow
steam); and (c) driving hollow steel piles. The soil inside the pile is then jetted out, after
which backfilling with sand is done. Figure 14.20 shows a schematic diagram of sand
drains. After backfilling the drill holes with sand, a surcharge is applied at the ground surface. The surcharge will increase the pore water pressure in the clay. The excess pore water
pressure in the clay will be dissipated by drainage—both vertically and radially to the sand
drains—thereby accelerating settlement of the clay layer. In Figure 14.20a, note that the
radius of the sand drains is rw . Figure 14.20b shows the plan of the layout of the sand drains.
The effective zone from which the radial drainage will be directed toward a given sand drain
is approximately cylindrical, with a diameter of de .
To determine the surcharge that needs to be applied at the ground surface and the
length of time that it has to be maintained, see Figure 14.16 and use the corresponding
equation, Eq. (14.15):
log B1 1
Uv,r 5
log b1 1
Ds(p)
r
sor
Ds(p)
r
sor
B1 1
R
Ds(f)
r
Ds(p)
r
(14.17)
Rr
The notations Ds(p)
r , sor , and Ds(f)
r are the same as those in Eq. (14.15); however, the lefthand side of Eq. (14.17) is the average degree of consolidation instead of the degree of
746 Chapter 14: Soil Improvement and Ground Modification
Surcharge
Groundwater
table
Sand
Vertical drainage
Sand drain
Hc
Clay
layer
Radial
drainage
Sand drain
radius rw
Radial
drainage
Vertical drainage
Sand
(a) Section
Sand drain
radius rw
de
(b) Plan
Figure 14.20 Sand drains
consolidation at midplane. Both radial and vertical drainage contribute to the average
degree of consolidation. If Uv,r can be determined for any time t2 (see Figure 14.16b), the
r 1 Ds(p)
r may be obtained easily from Figure 14.17. The procedure
total surcharge Ds(f)
for determining the average degree of consolidation (Uv,r ) follows:
For a given surcharge and duration, t2, the average degree of consolidation due to
drainage in the vertical and radial directions is
Uv,r 5 1 2 (1 2 Ur ) (1 2 Uv )
(14.18)
where
Ur 5 average degree of consolidation with radial drainage only
Uv 5 average degree of consolidation with vertical drainage only
The successful use of sand drains has been described in detail by Johnson (1970b).
As with precompression, constant field settlement observations may be necessary during
the period the surcharge is applied.
14.8 Sand Drains
747
Average Degree of Consolidation Due to Radial Drainage Only
Figure 14.21 shows a schematic diagram of a sand drain. In the figure, rw is the radius of the
sand drain and re 5 de>2 is the radius of the effective zone of drainage. It is also important
to realize that, during the installation of sand drains, a certain zone of clay surrounding them
is smeared, thereby changing the hydraulic conductivity of the clay. In the figure, rs is the
radial distance from the center of the sand drain to the farthest point of the smeared zone.
Now, for the average-degree-of-consolidation relationship, we will use the theory of equal
strain. Two cases may arise that relate to the nature of the application of surcharge, and they
are shown in Figure 14.22. (See the notations shown in Figure 14.16). Either (a) the entire
surcharge is applied instantaneously (see Figure 14.22a), or (b) the surcharge is applied in
the form of a ramp load (see Figure 14.22b). When the entire surcharge is applied instantaneously (Barron, 1948),
Ur 5 1 2 exp ¢
28Tr
≤
m
(14.19)
de
Smeared
zone
Clay
Hc
Sand
drain
rs
Surcharge
per unit area
re
rw
Figure 14.21 Schematic diagram of a sand drain
Surcharge
per unit area
␴(p) ␴(f)
Time
(a)
Figure 14.22 Nature of application of surcharge
␴(p) ␴(f)
Time
tc
(b)
748 Chapter 14: Soil Improvement and Ground Modification
where
m5
kh n2 2 S2
n2
n
3
S2
ln
¢
≤
2
1
1
¢
≤ ln S
S
4
ks
n2 2 S2
4n2
n2
(14.20)
in which
de
re
5
rw
2rw
rs
S5
rw
n5
(14.21)
(14.22)
and
kh 5 hydraulic conductivity of clay in the horizontal direction in the unsmeared zone
ks 5 horizontal hydraulic conductivity in the smeared zone
Cvrt2
(14.23)
Tr 5 nondimensional time factor for radial drainage only 5 2
de
Cvr 5 coefficient of consolidation for radial drainage
5
kh
(14.24)
De
B
Rg
Dsr(1 1 eav ) w
For a no-smear case, rs 5 rw and kh 5 ks, so S 5 1 and Eq. (14.20) becomes
m5 ¢
3n2 2 1
n2
≤
ln
(n)
2
n2 2 1
4n2
(14.25)
Table 14.5 gives the values of Ur for various values of Tr and n.
If the surcharge is applied in the form of a ramp and there is no smear, then
(Olson, 1977)
Tr 2
Ur 5
1
31 2 exp(2ATr )4
A
Trc
(for Tr < Trc )
(14.26)
and
Ur 5 1 2
1
3exp(ATrc ) 2 14exp(2ATrc )
ATrc
(for Tr > Trc )
(14.27)
14.8 Sand Drains
749
Table 14.5 Variation of Ur for Various Values of Tr and n, No-Smear Case [Eqs. (14.19) and (14.25)]
Degree of
consolidation Ur (%)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
Time factor Tr for value of n (5re , rw )
5
10
15
20
25
0
0.0012
0.0024
0.0036
0.0048
0.0060
0.0072
0.0085
0.0098
0.0110
0.0123
0.0136
0.0150
0.0163
0.0177
0.0190
0.0204
0.0218
0.0232
0.0247
0.0261
0.0276
0.0291
0.0306
0.0321
0.0337
0.0353
0.0368
0.0385
0.0401
0.0418
0.0434
0.0452
0.0469
0.0486
0.0504
0.0522
0.0541
0.0560
0.0579
0.0598
0.0618
0.0638
0.0658
0.0679
0
0.0020
0.0040
0.0060
0.0081
0.0101
0.0122
0.0143
0.0165
0.0186
0.0208
0.0230
0.0252
0.0275
0.0298
0.0321
0.0344
0.0368
0.0392
0.0416
0.0440
0.0465
0.0490
0.0516
0.0541
0.0568
0.0594
0.0621
0.0648
0.0676
0.0704
0.0732
0.0761
0.0790
0.0820
0.0850
0.0881
0.0912
0.0943
0.0975
0.1008
0.1041
0.1075
0.1109
0.1144
0
0.0025
0.0050
0.0075
0.0101
0.0126
0.0153
0.0179
0.0206
0.0232
0.0260
0.0287
0.0315
0.0343
0.0372
0.0401
0.0430
0.0459
0.0489
0.0519
0.0550
0.0581
0.0612
0.0644
0.0676
0.0709
0.0742
0.0776
0.0810
0.0844
0.0879
0.0914
0.0950
0.0987
0.1024
0.1062
0.1100
0.1139
0.1178
0.1218
0.1259
0.1300
0.1342
0.1385
0.1429
0
0.0028
0.0057
0.0086
0.0115
0.0145
0.0174
0.0205
0.0235
0.0266
0.0297
0.0328
0.0360
0.0392
0.0425
0.0458
0.0491
0.0525
0.0559
0.0594
0.0629
0.0664
0.0700
0.0736
0.0773
0.0811
0.0848
0.0887
0.0926
0.0965
0.1005
0.1045
0.1087
0.1128
0.1171
0.1214
0.1257
0.1302
0.1347
0.1393
0.1439
0.1487
0.1535
0.1584
0.1634
0
0.0031
0.0063
0.0094
0.0126
0.0159
0.0191
0.0225
0.0258
0.0292
0.0326
0.0360
0.0395
0.0431
0.0467
0.0503
0.0539
0.0576
0.0614
0.0652
0.0690
0.0729
0.0769
0.0808
0.0849
0.0890
0.0931
0.0973
0.1016
0.1059
0.1103
0.1148
0.1193
0.1239
0.1285
0.1332
0.1380
0.1429
0.1479
0.1529
0.1580
0.1632
0.1685
0.1739
0.1793
(continued)
750 Chapter 14: Soil Improvement and Ground Modification
Table 14.5 (continued)
Degree of
consolidation Ur (%)
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
Time factor Tr for value of n (5re , rw )
5
10
15
20
25
0.0700
0.0721
0.0743
0.0766
0.0788
0.0811
0.0835
0.0859
0.0884
0.0909
0.0935
0.0961
0.0988
0.1016
0.1044
0.1073
0.1102
0.1133
0.1164
0.1196
0.1229
0.1263
0.1298
0.1334
0.1371
0.1409
0.1449
0.1490
0.1533
0.1577
0.1623
0.1671
0.1720
0.1773
0.1827
0.1884
0.1944
0.2007
0.2074
0.2146
0.2221
0.2302
0.2388
0.2482
0.2584
0.2696
0.1180
0.1216
0.1253
0.1290
0.1329
0.1368
0.1407
0.1448
0.1490
0.1532
0.1575
0.1620
0.1665
0.1712
0.1759
0.1808
0.1858
0.1909
0.1962
0.2016
0.2071
0.2128
0.2187
0.2248
0.2311
0.2375
0.2442
0.2512
0.2583
0.2658
0.2735
0.2816
0.2900
0.2988
0.3079
0.3175
0.3277
0.3383
0.3496
0.3616
0.3743
0.3879
0.4025
0.4183
0.4355
0.4543
0.1473
0.1518
0.1564
0.1611
0.1659
0.1708
0.1758
0.1809
0.1860
0.1913
0.1968
0.2023
0.2080
0.2138
0.2197
0.2258
0.2320
0.2384
0.2450
0.2517
0.2587
0.2658
0.2732
0.2808
0.2886
0.2967
0.3050
0.3134
0.3226
0.3319
0.3416
0.3517
0.3621
0.3731
0.3846
0.3966
0.4090
0.4225
0.4366
0.4516
0.4675
0.4845
0.5027
0.5225
0.5439
0.5674
0.1684
0.1736
0.1789
0.1842
0.1897
0.1953
0.2020
0.2068
0.2127
0.2188
0.2250
0.2313
0.2378
0.2444
0.2512
0.2582
0.2653
0.2726
0.2801
0.2878
0.2958
0.3039
0.3124
0.3210
0.3300
0.3392
0.3488
0.3586
0.3689
0.3795
0.3906
0.4021
0.4141
0.4266
0.4397
0.4534
0.4679
0.4831
0.4992
0.5163
0.5345
0.5539
0.5748
0.5974
0.6219
0.6487
0.1849
0.1906
0.1964
0.2023
0.2083
0.2144
0.2206
0.2270
0.2335
0.2402
0.2470
0.2539
0.2610
0.2683
0.2758
0.2834
0.2912
0.2993
0.3075
0.3160
0.3247
0.3337
0.3429
0.3524
0.3623
0.3724
0.3829
0.3937
0.4050
0.4167
0.4288
0.4414
0.4546
0.4683
0.4827
0.4978
0.5137
0.5304
0.5481
0.5668
0.5868
0.6081
0.6311
0.6558
0.6827
0.7122
14.8 Sand Drains
751
Table 14.5 (continued)
Time factor Tr for value of n (5re , rw )
Degree of
consolidation Ur (%)
91
92
93
94
95
96
97
98
99
5
10
15
20
25
0.2819
0.2957
0.3113
0.3293
0.3507
0.3768
0.4105
0.4580
0.5391
0.4751
0.4983
0.5247
0.5551
0.5910
0.6351
0.6918
0.7718
0.9086
0.5933
0.6224
0.6553
0.6932
0.7382
0.7932
0.8640
0.9640
1.1347
0.6784
0.7116
0.7492
0.7927
0.8440
0.9069
0.9879
1.1022
1.2974
0.7448
0.7812
0.8225
0.8702
0.9266
0.9956
1.0846
1.2100
1.4244
where
Trc 5
Cvr tc
d2e
(see Figure 14.22b for the definition of tc )
(14.28)
and
A5
2
m
(14.29)
Average Degree of Consolidation Due to Vertical Drainage Only
Using Figure 14.22a, for instantaneous application of a surcharge, we may obtain the
average degree of consolidation due to vertical drainage only from Eqs. (1.74) and (1.75).
We have
Tv 5
p Uv (%) 2
B
R
4
100
(for Uv 5 0 to 60%)
[Eq. (1.74)]
and
Tv 5 1.781 2 0.933 log 3100 2 Uv (%)4
(for Uv . 60%)
[Eq. (1.75)]
where Uv 5 average degree of consolidation due to vertical drainage only, and
Tv 5
Cvt2
H2
where Cv 5 coefficient of consolidation for vertical drainage.
[Eq. (1.69)]
752 Chapter 14: Soil Improvement and Ground Modification
For the case of ramp loading, as shown in Figure 14.22b, the variation of Uv with Tv
can be expressed as (Olson, 1977):
For Tv ; Tc:
Uv 5
Tv
2
1
e1 2
S 4 31 2 exp(2M 2Tv )4 f
Tc
Tv M
(14.30)
For Tv ; Tc:
Uv 5 1 2
2
1
S 4 3exp(2M 2Tc ) 2 14exp(2M 2Tv )
Tc M
(14.31)
where
p
(2mr 1 1)
2
m r 5 0, 1, 2, c
M5
Tc 5
Cvtc
(14.32)
H2
where H 5 length of maximum vertical drainage path. Figure 14.23 shows the variation
of Uv (%) with Tc and Tv.
0
10
0.04 0.1 0.2
Tc 0
Degree of Consolidation, Uv (%)
20
30
0.5
1.0
5.0
2.0
40
50
60
70
80
90
100
0.01
0.1
1.0
Time Factor, Tv
Figure 14.23 Variation of Uv with Tv and Tc [Eqs. (14.30) and (14.31)]
10
14.8 Sand Drains
753
Example 14.2
Redo Example 14.1, with the addition of some sand drains. Assume that
rw 5 0.1 m, de 5 3 m, Cv 5 Cvr , and the surcharge is applied instantaneously. (See
Figure 14.22a.) Also assume that this is a no-smear case.
Solution
Part a
The total primary consolidation settlement will be 167.7 mm, as before.
Part b
From Example 14.1, Tv 5 0.36. Using Eq. (1.74), we obtain
Tv 5
p Uv (%) 2
B
R
4
100
or
Uv 5
4Tv
(4) (0.36)
3 100 5
3 100 5 67.7%
p
Å p
Å
Also,
n5
de
3
5
5 15
2rw
2 3 0.1
Again,
Tr 5
Cvr t2
d2e
5
(0.36) (9)
(3) 2
5 0.36
From Table 14.5 for n 5 15 and Tr 5 0.36, the value of Ur is about 77%, Hence,
Uv,r 5 1 2 (1 2 Uv ) (1 2 Ur ) 5 1 2 (1 2 0.67) (1 2 0.77)
5 0.924 5 92.4%
Now, from Figure 14.17, for Dspr >sor 5 0.548 and Uv,r 5 92.4%, the value of
Dsfr >Dspr < 0.12. Hence,
Ds(f)
r 5 (115) (0.12) 5 13.8 kN , m2
■
754 Chapter 14: Soil Improvement and Ground Modification
Example 14.3
Suppose that, for the sand drain project of Figure 14.20, the clay is normally consolidated. We are given the following data:
Clay: Hc 5 4.57 m (two-way drainage)
Cc 5 0.31
eo 5 1.1
Effective overburden pressure at the middle of the clay layer
5 47.92 kN>m2
Cv 5 106.15 3 10 24 m2>day
Sand drain: rw 5 0.091 m
de 5 1.83 m
Cv 5 Cvr
A surcharge is applied as shown in Figure 14.24. Assume this to be a no-smear case.
Calculate the degree of consolidation 30 days after the surcharge is first applied. Also,
determine the consolidation settlement at that time due to the surcharge.
Solution
From Eq. (14.32),
Tc 5
Cv tc
H2
5
(106.15 3 10 24 m2>day) (60)
4.57 2
¢
≤
2
5 0.122
and
Tv 5
Cvt2
H2
5
(106.15 3 10 24 ) (30)
4.57 2
¢
≤
2
5 0.061
Surcharge (kN/m2)
95.84 kN/m2 ␴(p) ␴( f )
60 days tc
Time
Figure 14.24 Ramp load for a sand
drain project
14.8 Sand Drains
755
Using Figure 14.23 for Tc 5 0.123 and Tv 5 0.061, we have Uv < 9%. For the sand drain,
n5
de
1.83
5
5 10
2rw
(2) (0.091)
From Eq. (14.28),
Trc 5
Cvr tc
(106.15 3 10 24 ) (60)
5
d2e
5 0.19
(1.83) 2
and
Tr 5
Cvr t2
d2e
5
(106.15 3 10 24 ) (30)
(1.83) 2
5 0.095
Again, from Eq. (14.26),
Tr 2
Ur 5
1
31 2 exp(2ATr ) 4
A
Trc
Also, for the no-smear case,
m5
3(10) 2 2 1
n2
3n2 2 1
102
ln
(n)
2
5
ln
(10)
2
5 1.578
n2 2 1
4n2
102 2 1
4(10) 2
and
A5
2
2
5
5 1.267
m
1.578
so
0.096 2
Ur 5
1
31 2 exp(21.267 3 0.095) 4
1.267
5 0.03 5 3%
0.19
From Eq. (14.18),
Uv,r 5 1 2 (1 2 Ur ) (1 2 Uv ) 5 1 2 (1 2 0.03) (1 2 0.09) 5 0.117 5 11.7%
The total primary settlement is thus
Sc(p) 5
5
sor 1 Ds(p)
r 1 Dsfr
CcHc
log B
R
1 1 eo
sor
(0.31) (4.57)
47.92 1 95.84
log ¢
≤ 5 0.332 m
1 1 1.1
47.92
and the settlement after 30 days is
Sc(p)Uv,r 5 (0.322) (0.117) (1000) 5 37.67 mm
■
756 Chapter 14: Soil Improvement and Ground Modification
14.9
Prefabricated Vertical Drains
Prefabricated vertical drains (PVDs), also referred to as wick or strip drains, were originally
developed as a substitute for the commonly used sand drain. With the advent of materials science, these drains began to be manufactured from synthetic polymers such as polypropylene
and high-density polyethylene. PVDs are normally manufactured with a corrugated or channeled synthetic core enclosed by a geotextile filter, as shown schematically in Figure 14.25.
Installation rates reported in the literature are on the order of 0.1 to 0.3 m>s, excluding
equipment mobilization and setup time. PVDs have been used extensively in the past for
expedient consolidation of low-permeability soils under surface surcharge. The main advantage of PVDs over sand drains is that they do not require drilling; thus, installation is much
faster. Figures 14.26a and b are photographs of the installation of PVDs in the field.
Design of PVDs
The relationships for the average degree of consolidation due to radial drainage into sand
drains are given in Eqs. (14.19) through (14.24) for equal-strain cases. Yeung (1997) used
these relationships to develop design curves for PVDs. The theoretical developments used
by Yeung are given next.
Figure 14.27 shows the layout of a square-grid pattern of prefabricated vertical
drains. See also Figure 14.25 for the definition of a and b). The equivalent diameter of a
PVD can be given as
dw 5
2(a 1 b)
p
(14.33)
Now, Eq. (14.19) can be rewritten as
Ur 5 1 2 exp ¢2
8Cvr t d 2w
d2w
d2e m
≤ 5 1 2 exp ¢2
8Trr
≤
ar
b
Polypropylene
core
Geotextile
fabric
a
Figure 14.25 Prefabricated vertical
drain (PVD)
(14.34)
14.9 Prefabricated Vertical Drains
(a)
(b)
Figure 14.26 Installation of PVDs in the field (Note: (b) is a closeup view of (a))
(Courtesy of E. C. Shin, University of Incheon, Korea)
757
758 Chapter 14: Soil Improvement and Ground Modification
d
d
a
b
Dia. dw
Smear zone
Diameter ds
Diameter de
Figure 14.27 Square-grid pattern layout of prefabricated vertical drains
where de 5 diameter of the effective zone of drainage 5 2re. Also,
Trr 5
Cvr t
d2w
4
kh
3n2 2 S2
n
n
≤
2
¢
≤ 1 (n2 2 S2 ) ln S
ar 5 n2m 5 2
ln
¢
2
S
4
ks
n 2S
(14.35)
(14.36)
and
n5
de
dw
(14.37)
From Eq. (14.34),
Trr 5 2
ar
ln (1 2 Ur )
8
or
Trr
ln (1 2 Ur )
52
(14.38)
ar
8
Table 14.6 gives the variation of (Trr)1 with Ur. Also, Figure 14.28 shows plots of
a r versus n for kh>ks 5 5 and 10 and S 5 2 and 3.
(Trr ) 1 5
14.9 Prefabricated Vertical Drains
Table 14.6 Variation of (Trr ) 1 with Ur [Eq. (14.38)]
Ur (%)
(Trr )1
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
0
0.006
0.013
0.020
0.028
0.036
0.045
0.054
0.064
0.075
0.087
0.100
0.115
0.131
0.150
0.173
0.201
0.237
0.288
0.374
100,000
100,000
S3
S2
2
kh /ks 10
10,000
3
kh /ks 10
10,000
5
5
␣
␣
S1
S1
1,000
1,000
100
100
0
10
20
30
n
40
50
60
0
(a)
Figure 14.28 Plot of ar versus n: (a) S 5 2: (b) S 5 3 [Eq. (14.36)]
10
20
30
n
(b)
40
50
60
759
760 Chapter 14: Soil Improvement and Ground Modification
Following is a step-by-step procedure for the design of prefabricated vertical drains:
Step 1. Determine time t2 available for the consolidation process and the Uv,r
required therefore [Eq. (14.17)]
Step 2. Determine Uv at time t2 due to vertical drainage. From Eq. (14.18)
Ur 5 1 2
Step 3.
Step 4.
Step 5.
Step 6.
1 2 Uv,r
1 2 Uv
(14.39)
For the PVD that is to be used, calculate dw from Eq. (14.33).
Determine (Trr ) 1 from Eqs. (14.38) and (14.39).
Determine Trr from Eq. (14.35).
Determine
ar 5
Trr
.
(Trr ) 1
Step 7. Using Figure 14.28 and ar determined from Step 6, determine n.
Step 8. From Eq. (14.37),
de 5 n
c
Step 7
dw
c
Step 3
Step 9. Choose the drain spacing:
de
1.05
de
d5
1.128
d5
(for triangular pattern)
(for square pattern)
A Case History
The installation of PVDs combined with preloading is an efficient way to gain strength in soft
clays for construction of foundations. An example of a field study can be found in the works
of Shibuya and Hanh (2001) which describes a full-scale test embankment 40 m 3 40 m
in plan constructed over a soft clay layer located at Nong Ngu Hao, Thailand. PVDs were
installed in the soft clay layer in a triangular pattern (Figure 14.29a). Figure 14.29b shows the
pattern of preloading at the site along with the settlement-time plot at the ground surface
below the center of the test embankment. Maximum settlement was reached after about four
months. The variation of the undrained shear strength (cu) with depth in the soft clay layer
before and after the soil improvement is shown in Figure 14.29c. The variation of cu with
depth is based on field vane shear tests. The undrained shear strength increases by about 50 to
100% at various depths.
14.10
Lime Stabilization
As mentioned in Section 14.1, admixtures are occasionally used to stabilize soils in the field—
particularly fine-grained soils. The most common admixtures are lime, cement, and lime–fly
ash. The main purposes of stabilizing the soil are to (a) modify the soil, (b) expedite
construction, and (c) improve the strength and durability of the soil.
14.10 Lime Stabilization
761
Sand mat
1.7 m
Fill
0.8 m
12 m
40 m
PVD 1.0 m spacing
triangular pattern
(a)
5
Undrained shear strength, cu (kN/m2)
10
20
30
0
2.0
4
1.0
Depth (m)
Test embankment height (m)
3.0
0.5
After
improvement
8
Before
improvement
0
12
Settlement (m)
0.2
16
0.4
(c)
0.6
0.8
1.0
0
40
80
120
Time (days)
(b)
160
200
Figure 14.29 Shibuya and Hanh (2001) study
of a full-scale test embankment at Nong Ngu Hao
(Thailand): (a) test embankment; (b) test embankment height and ground settlement with time;
(c) undraine